L. B. CLEVELAND 




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NOTES AND EXAMPLES 



IN — 



MECHANICS; 



WITH AN APPENDIX ON THE 



GRAPHICAL STATICS OF MECHANISM ; 



IRVING P. CHURCH, C. E., 

Professor of Applied Mechanics and Hydraulics , College of Civil Engineering, 
Cornell University. 



SECOND EDITION, REVISED AND ENI^ARGED. 



NEW YORK : 

JOHN WILEY & SONS. 

London : CHAPMAN & HALL, Limited. 

1903. 






Copyright, 1891, 




BY 


Irving 


P. Church. 




G ' -f-t 


. i r> 


;• y ■-- . r i. ^ 


Oct. 


31 1935 



ERRATA. 
P. 63. In Fig. 76 for " 12.2' " read " 12.5'." 
P. 69. I^ast line but one, for "4937 lbs." read "4973 lbs." 
P. 70. Second line, for " 7063 lbs." read " 7027 lbs." 
Plate V of Appendix ; in Fig. 17 [A] for " S " read " S2 " 



3^ - 3/^c^ ^ 



PKEFACE. 



The following pages form a companion volume to tlie writer's Mechanics of 
Engineering, and contain various notes and many practical examples, botli 
algebraic and numerical, serving to illustrate more fully the application of 
fundamental principles in Mechanics of Solids ; together with a few paragraphs 
relating to the Mechanics of Materials, and an Appendix on the "Graphical 
Statics of Mechanism " 

Advantage has been taken of the use of preliminary impressions m the 
classroom to make corrections in the electrotype plates; and it is therefore 
thought that the present complete edition is comparatively free from typo- 
graphical errors. 

In the Appendix are presented many of the problems of Prof. Herrmann s 
^'Zur Graphischen Staiik der Maschinengeiriebe " in what seems to the writer 
a clearer form than in the original (for reasons stated on the first page of 
Appendix). In this part of the work, the text and diagrams not being adjacent, 
alternate pages have been left blank in such a way that any diagram and its 
appropriate text can be kept in view simultaneously. 

Besides his indebtedness to Prof. Herrmann's work, the writer would grate- 
fully acknowledge the kindness of .the Messrs. Wiley in securing a higher 
order of excellence in the execution of the diagrams than had at first been 
contemplated. 

In references to the writer's Mechanics of Engineering the abbreviation 

M. of E. is used. 

Cornell University, 
Ithaca, N. Y., March, 1892. 



PEEFACE TO SECOND EDITION. 



For this second edition the plates of the first have been carefully revised 
and corrected (except as indicated in the errata on the opposite page), and an 
entirely new chapter added (Chap. VIII, pp. 119-133), containing various notes 
and explanations, as also many examples for practice. 



Ithaca, January, 1897. 



CONTENTS. 



CHAP. I. Definitions. Principles. Center of Geavitt. 

PAGES 

^§ 1-15. Illustrations of Forces. Mass, Weight, Equilibrium. Fundamental 

Theorem of the Integral Calculus. Centers of Gravity. Simpson's Kule. 1-14 

CHAP. II. Pkinciples and Problems Involving Non- 
Concurrent Forces in a Plane. 

5§ 16-40a. Conditions of Equilibrium. Classification of Kigid Bodies. Two- 
force Pieces ; Three-force Pieces, etc. Levers ; Bell-crank ; Cranes, 
Simple and Compound. Kedundant Support. Two Links ; Rod and 
Tumbler ; Door; Wedge and Block. Roof Truss, and Cantilever Frame. 15-42 

CHAP. III. Motion of a Material Point. 

5§ 41-51. Velocity. Acceleration. Momentum. Cord and Weights, Lifting 
a Weight. Harmonic Motion. Ballistic Pendulum. Balls and Spring. 
Cannon as Pendulum. Simple Circular Pendulum. .'.'^..VJ 43-53 

CHAP. IV. Numerical Examples in Statics of Rigid Bodies 
AND Dynamics of a Material Point. 

)§ 52-72a. Center of Gravity. Toggle Joint. Crane. Door. Roof Truss. 
Train Resistance. Motion on Inclined Plane. Block Sliding on Circular 
Guide, etc. Harmonic Motion of Piston. Conical Pendulum. Weighted 
Governor. Motion in Curve. Motion of Weighted Piston with Steam 
used Expansively. Ball Falling on Spring 54-76 

CHAP. V. Moment of Inertia of Plane Figures. 

)§ 72b-76. Section of I-beam ; of Box-beam. Irregular Figures, by Simp- 
son's Rule. Graphical Method 77-82 

CHAP. VI. Dynamics of a Rigid Body. 

}§ 77-97. Rotary Motion. Pendulum. Speed of Fly-wheel. Centrifugal 
Action on Bearings. " Centrifugal Couple." Piles, Kinetic Energy of 
Rotary Motion. Work and Energy. Numerical Examples. Action of 
Forces in Locomotive. The Appold and Carpentier Dynamometers. 
Boat Rowing. Solutions of Numerical Examples. Work of Rolling 
Resistance. Strap Friction ; Examples 83-106 

CHAP. VII. Mechanics of Materials and Graphical Statics. 

5§ 98-107. Stresses in a Rod in Tension. Rivet-spacing in a Built Beam. 
I-beams; without using Moment of Inertia. " Incipieiit Flexure "in 
a Column. Tests of Wooden Posts. The Pencoyd Experiments in 
Columns. Graphical Constructions 107-118 

CHAP. VIII. Miscellaneous Notes. 

5§ 108-116. Center of Gravity. The Time-velocity Curve. Reduction of 
Moment of Inertia of Plane Figure. Miscellaneous Examples. The 
" Imaginary System " in Motion of Rigid Body. Angular Motion 119-133 

APPENDIX ON THE Graphical Statics of Mechanism. 1-34 

(See p. 28 of the Appendix for its Table of Contents.) 



NOTES AND EXAMPLES IN MECHANICS. 



CHAPTEK I. 

Definitions. Principles. Centre of Gravity. 

1. Applied Mechanics is perhaps a more common term for the 
same thing than " Mechanics of Engineering." " Pure Mechan- 
ics" is another name for Analytical Mechanics, which deals with 
the subject entirely from a mathematical point of view. 

2. Abstract Numbers. — In experimental investigations in which 
formulae are to be deduced, it is best to throw experimental 
coefficients into the form of abstract numbers, if possible, for 
these are immediately comparable with those of another experi- 
menter in the same field, if the latter follows the same plan^ 
whether he uses the same units for space, force, and time, or not. 

Thus : if the coefficient of friction be defined as the ratio of 
the friction [force] to the normal pressure [force] producing it, 
we obtain the same number for it in a definite experiment, 
whether we express our forces in pounds or in kilograms. 

3. Forces. — One of the most important things to be acquired 
in dealing with the practical problems of this study is a proper 
conception of forces. We do not use the word force in any 
abstract general sense, nor in any popular sense, such as is in- 
stanced in the E'ote of § 15c, M. of E. It should always mean 
the pull, pressure, rub, attraction (or repulsion), of one body upon 
another, and always implies the existence of a simultaneous, equal, 
and opposite force exerted by that other body on the first body, 
i.e., the reaction ; but this reaction will not come up for consid- 
eration in any problem unless this "first" body is under treat' 



x^ NOTES AND EXAMPLES IN MECHANICS. 

ment as regards the forces acting on it. In most problems in 
Mechanics we have one or more definite rigid bodies under con- 
sideration, one at a time, in whose treatment we must form clear 
conceptions of tlie forces acting on it ; and these always emanate 
from other bodies. 

Hence in no case should we call anything a force unless we 
can conceive of it as capable of measurement by a spring-balance, 
and are aUe to say from what other tody it coines. 

For example^ a body said to weigh 30 lbs. lies at rest on a 
smooth level table, which is the only body with which it is in 
contact. When considered by itself this body is acted on by only 
two other bodies in a manner which justifies the use of the- word 
force ; viz., the action of the earth upon it is a vertical downward 
attraction (force) of 30 lbs. ; while the action of the table upon it 
is an upward pressure (force) of 30 lbs. (We here ignore the 
atmosphere whose pressures on the body are balanced in every 
direction.) But suppose the same body and the table with which 
it is in contact to be allowed to fall, from rest, in a vacuum. The 
two bodies, during the fall, remain apparently in as close contact 
as before ; but now the upper body is under the action of only 
one force, viz., the downward attraction of the earth, 30 lbs. ; and 
there is no pressure of the upper body against the table, and con- 
sequently no pressure of the table against the upper body. 

As another instance, an iron rod rests horizontally on two 
level-faced supports, at its extremities, and bears a load of 60 lbs. 
in the middle. When this rod is considered "/r^^," i.e., when those 
other bodies which act on it in a ^\force-2ih\Q^^ way are supposed 
removed (their places being for present purposes taken by the 
respective forces with Avhich they act on the first body), we find 
it to be under the action of four forces, viz. : a pressure on its 
middle, vertical and downward, of 60 lbs. from its load ; the 
downward attraction of the earth on it, i.e., its own weight, say 
10 lbs. (which is really distributed among all of its particles, but 
which, so far as the equilibrium, or state of rest or motion, of the 
body is concerned, is the same as if applied at the centre of 
gravity, viz., the middle of the rod) ; and the two upward press- 
ures of the two supports against the ends of the rod, these being 



FOECES. 



35 lbs. each. If the nature of the investigation requires it, we 
maj go on and consider one of the supports bj itself, or " free" ; 
in which case, whatever the actions of other bodies on it may be, 
that of the rod will be a downward force of 35 lbs., the equal 
and opposite of the 35 lbs. upward pressure of the support against 
the rod. These pressures of the two supports against the rod are 
usually called the ''Reactions of the Supports." 

As another instance : a ball of 10 lbs. weight hangs at rest by 
a cord attached to a support above. The cord is of course ver- 
tical. This ball is under the action of two forces, viz., a down- 
ward attraction of 10 lbs. emanating from the earth, and an 
upward pull of 10 lbs. emanating from the cord. 

A portion of the above cord, taken in the part under ten- 
sion, is under the action of two forces, thus : the part just 
above it exerts an upward pull of 10 lbs. upon it, and that below 
it exerts a downward pull of 10 lbs. upon it. (We here neglect 
the weight of the portion of cord considered as presumably very 
small.) In such a case the tension of the cord is said to be 10 
lbs. (not 20 lbs.). 

Further illustration. Fig. 1 shows a prismatic rod CB lean- 
ing against the smooth vertical side of a block. Both rest on a 
rough horizontal plane. The rod is under 
three forces, viz. : its weight G acting 
vertically downwards through its middle ; 
the pressure of the wall against it, P 
(which, since the wall-surface is perfectly 
smooth, must be horizontal and points 
toward the right) ; and a third force, Q^ 
the pressure of the floor against the rod. 
Since the rod and block are at rest, P and 
G intersecting at A, it must be that the floor is sufficiently rough 
to enable the pressure Q to deviate from the vertical (that is, from 
the normal to plane of floor) by as much as the angle AB V, at 
least; for, as will be proved later, if three forces act on a body 
and it remains at rest, the three lines of action must intersect in 
a common point. 

We next consider the block, or wall, by itself, and find it to 




Fig. 1. 



4 NOTES AND EXAMPLES IN MECHANICS. 

be under the action of P\ the equal and opposite of P^ and 
therefore pointing horizontally toward the left ; of G\ its weight; 
and a pressure, R^ from the floor, whose line of action is deter- 
mined by the fact that it must be the same line of action as that of 
the (ideal) resultant of the forces G' and P\ since if three forces 
balance, i.e., are in equilibrium, any one of them must be the 
equal and opposite of the resultant of the other two and have the 
same action-line. Forming a parallelogram, therefore, on the 
forces P' and G\ first conceiving each of the two forces to be 
transferred in its line of action to their point of intersection, A\ 
the diagonal of this parallelogram represents the equal and oppo- 
site of P^ and has the same line of action. 

If this diagonal intersects the floor on the left of the lower 
left-hand corner of the block, the supposed stability is impossible, 
unless the block is cemented to the floor. 

4. Mass and Weight. — The question of mass will be further 
discussed in a subsequent chapter, p. 53, M. of E. By weight we 
are always to understand the force of the attraction which the 
earth exerts upon the body, and not the amount of matter (mass) 
in it. This weight will therefore be different in different latitudes 
and at different distances from the centre of the earth, and re- 
quires a spring -lalance for its determination. Physicists also use 
the word weight in this sense (force). 

6. The Heaviness of a body, in the sense used in this study, is 
something quite different from its total weight. For instance, if 
the substance of the body is not uniform in composition and den- 
sity, we cannot speak of the heaviness of the body as a whole, 
since its various portions have not a common heaviness ; however, 
we may speak of its average heaviness, which might be of some 
use in certain problems, and would be the quotient of its total 
weight divided by its volume. 

Since heaviness is not an abstract number, it would not be 
sufficient to say that a certain substance has a heaviness of 40, 
for instance, nor even 40 lbs. ; the full statement must be 40 lbs. 
per cubic foot ; which is equivalent to the statement that the 
heaviness is 0.540 ton per cubic yard. 

6, Rigid Body. — As an illustration of the definition in § 10, 



EQUILIBRIUM — TRANSMISSIBILITY OF FORCE. 5 

M. of E., it may be said that if a horizontal bar, supported at its 
extremities, is so moderately loaded that the deflection or sinking 
of the central point is only about one 3-hundredth part, for in- 
stance, of the span or distance between the supports, it is suffi- 
ciently accurate, for most purposes, to consider that there has been 
no change in the length of the horizontal projection of any dis- 
tance measured along the bar. (Where such a consideration is 
inadmissible, attention will be called to it.) 

7. Equilibrium. — Besides speaking of a system of forces being 
in equilibrium, the phraseology is also sometimes used that the 
rigid hody is in eqidlibvmm under the forces acting on it (§ 40<:t). 

The reservation made in § 11, M. of E., as to state of motion 
refers to the fact that any alteration in the distribution of forces 
acting on a rigid body will usually cause a difference in the inter- 
nal strains and stresses produced, though the state of motion may 
or may not be affected, according as any second 

system of forces applied to the body, on removal of the first, has 
a different resultant from that of the first, or the same resultant. 

8. Division of the Subject. — As to the division given in § 12, 
M. of E., Sir William Thomson, the noted English physicist, has 
adopted a different nomenclature, which is getting into wider and 
wider use. He makes the term Dynamics include both statics 
and dynamics (i.e., what is here and by Rankine and Continental 
writers called Dynamics), and replaces their word Dynamics by 
Kinetics. 

9. Transmissibility of Force. Resultant. — The principle of the 
transmissibility of force refers only to the state of motion of the 
rigid body. For instance (see Fig. 2), as far as the rest or motion 
of the sickle-shaped body is concerned, it is y/ 
immaterial whether the force P balance P' 
(being equal and opposite to it and in the 
same line) by being applied at 0, or by being 
applied at A ; but in the former case the part y^ 
ABO would be under a bending strain, and in fig. s. 

the latter would be under no strain. 

It must be remembered that the resultant of a given system 
of forces is always a purely imaginary force ; that is, all we mean 




6 NOTES AND EXAMPLES IN MECHANICS. 

is this : that if the given forces were removed and their resultant 
acted in their stead, in jprojper position as well as magnitude, the 
state of motion of the rigid body would not be any different from 
what it would be without the replacement. 

10. Parallelogram of Forces; or Triangle of Forces. — By some 
it is contended that this geometrical construction, or relation, the 
Parallelogram of Forces, should give place to the triangle of 
forces ; i.e., that the resultant (a line laid off to scale representing it) 
is equal to the third side of a triangle whose other two sides are the 
two given forces ; but a construction of that nature does not show 
the resultant as acting through the same point as the two com- 
ponents, which should be the case if the construction is to give 
the position, as well as the magnitude of the resultant. 

It is true that the systematic application of the triangle of 
forces gives rise to the methods of Graphical Statics, as will be 
seen, but in that case the magnitudes of all the forces (or rather 
lines of definite length representing them and parallel to them) 
are drawn on a separate part of the paper from that containing 
the figure of the body acted on. 

11. General Remarks on Forces. — It is not such a simple 
matter as it might at first appear, to bring to bear upon a given 
body a force of prescribed magnitude and direction at a specified 
point. For example, if we have the idea that a given tension can 
be ]3roduced in a vertical cord by hanging upon it a body whose 
w^eight is the given amount, we must remember that the tension 
in the cord will be equal to the body's w^eight only in case the 
hody is at rest or moving in a right line with unchanging velocity 
(i.e., describing equal spaces in equal times). 

While we can always be sure that the weight of the body itself 
(or action of the earth on it) is €i force of constant value acting 
on it at all times and in a vertical downward direction, the values 
of the pressures or pulls exerted upon it by neighboring bodies 
depend on the state of motion of the bodies concerned, as well as 
on their weights. For instance, suppose that w^e wish to observe 
the effect, upon a block of metal placed on a smooth level table 
and weighing 32.2 lbs., of bringing to bear upon it a horizontal 
force of 10 lbs. If (see Fig. 3) we attach to it a light flexible 



THEOKEM OF THE INTEGRAL CALCULUS. 




Fig. 3. 



(but inextensible) cord, led off horizontally to the right and then 

passing over a frictionless and very light 

pulley, M, and fasten a 10-lb. weight, 

B, to the end of the right-hand vertical 

portion, we find, by interposing a light 

spring-balance between the two parts of 

the cord at A^ that when motion is 

allowed to begin (the cord being pulled 

straight before the right-hand weight is allowed to sink freely), 

the tension at A is not 10 lbs., but only 7.63 lbs. 

If now we gradually increase the weight at B in successive 
experiments until the spring-balance in the cord at A shows a 
reading of just 10 lbs. (the desired force), we find that the weight 
at B has reached a value of 14.5 lbs. (The reasons for this will 
appear in Ex. 2 of § 57, M. of E.) 

The effect of the 10-lb. force on the block is then noted to be 
as follows : The block begins to move toward the right (having 
been initially at rest) with an accelerated motion. In the first 
second it passes over 5 ft. ; in the second second, 15 ft. ; in the 
third, 25 ft. ; and so on, as the odd numbers, 1, 3, 5, etc. In other 
words, the total distance from the start is equal to 5 ft. X the 
square of the total time in seconds. 

12. Review of the Fundamental Theorem of the Integral Cal- 
culus. — This must be thoroughly reviewed and understood before 
using the integral calculus in the theory of the centre of gravity, 

or elsewhere in Mechanics. The 
problems to which we apply the 
integral calculus will almost always 
be of a nature calling for the sum- 
ming up of a vast number of very 
small quantities all of which are 
alike in form and usually consist 
of the product of two or more fac- 
tors, one of which [the differential'] 
is very small, from which it comes 
that the product itself is small. (This is the most available form 
in which to conceive of the operation, but, as will be seen, it is 




•^i^a;- 



FiG. 4. 



8 NOTES AND EXAMPLES IN MECHANICS. 

after all a fictitious description of what is done, being put 
into that shape to avoid the continued repetition of a very 
lengthy phraseology.) This theorem and operation are most 
readily appreciated by the aid of graphic representation, as 
follows : 

Let ON be any portion of any algebraic curve, its equation 
being 

y —func. of X ; or, y — (p(x), .... (1) 

At any point m of the curve there is a special value of a? and 
of y, and also of the angle a^ which the tangent-line there drawn 
makes with the axis X. 

From the Differential Calculus we know that tan a is the " first 
differential coefficient " of y with respect to x (or " first deriva- 
tive," or simply " derivative"), and may be written (p\x) ; whence 
with symbols, 

dy ,, . 
tan a = -^; or tan a = (p\x) (2) 

Now divide the distance A . . . B into a large number of 
parts, not necessarily equal, the length of any one being called 
z/a?, and raise an ordinate at each point of subdivision. Where 
each such ordinate cuts the curve, as at m, draw a tangent line to 
the curve, and a horizontal line as m . . . h. These two lines in- 
tercept a small length a . . .h on the next ordinate on the right. 
All such intercepts are shown drawn in heavy lines, ah is typical 
of any one of these intercepts. From the right triangle concerned 
we now have 

ah = mh . tan a; i.e., ah = (tan a)^x = [^(p\x)']/}x. . (3) 

Project all the lengths like ah on a convenient vertical line, 
as ^D, and note that their sum is, of course, not quite equal to 
J^I), or NO, which is yn—Vo- This sum we may express as 
2^ \^(p\x)]dx ; and hence state that 

^^[0'(cc)] Jaj is almost eqical to y^ — y^* . . (4) 



THEOREM OF THE INTEGRAL CALCULUS. 9 

Or, since yn — y^ i^iay be written [0(a?)Jx = a:n — \_^{^)\x=x^^ 
(p{x), we may re- state the fact thus : 

^^[(p\x)']/lxal77iost = \ (p{x). . , . , (5) 

La-0 

Since it is evidently a geometrical possibility to increase the 
number of ^cc's between A and JB until the discrepancy between. 

^^\(^\x)']/lx and P''0(a^) is less than any value that maybe as- 
signed, we may say that "^ (p{x) is the lirait which ^^Tcp^x^J^x 

\—Xo 

approaches as the subdivision becomes finer and finer. This limit 
may also be expressed by the notation / "^ r^\xy\dx, and we 
accordingly write 

which limit = 0(a?). . (6) 

The utility of all this is that if in any problem we note that 
the sum of a series of terms, each one of which is the product of 
a small portion, Ax, of the axis of Xby a function of x (the same 
funclion in all) {Ax being the difference between the value of x 
for any term and that for the next term) is an approximation to 
the desired result; and if, furthermore, the desired result is the 
limit approached in value by the sum of the series as the sub- 
division becomes finer and finer along the axis of X, then the 
desired result will be obtained as follows : Find the anti-deriva- 
tive of the ahove function of x, and substitute in it first, for a?, the 
value, a?^ , which x assumes at the upper end of the series ; and 
secondly, substitute the value x^ of x at the lower end of the 
series. The difference of the expressions so obtained is the desired 
result. (IN'oTE. — By anti-derivative we are to understand that 
function of cc, the differentiation of which will give rise to the 
given function. Thus, the anti-derivative of 0X^) = ^^ ^^ 
<p{x) = ^x^ 4" const. ; and in applying the above rule we may 




10 NOTES AND EXAMPLES IN MECHANICS. 

omit the constant, knowing that it would cancel out in the sub- 
traction indicated.) 

13. Example of the Foregoing. — We wish to find by calculus 
the total moment of the homogeneous paraboloid of revolution in 
Fig. 5 about the axis Y\ the axis of the 
solid being horizontal and coinciding wnth 
the axis X. (By ''moment" of a small 
"^ weight we mean the product of the weight 
by the length of the perpendicular let fall 
upon it from a given line or plane). The 
Fig. 5. soHd is bouudcd on the right by a circular 

base, KN^ perpendicular to axis X. Conceiving the solid to be 
divided into a great number of circular disks, or laminge, perpen- 
dicular to axis X^ all of the same thickness, z/^, but of different 
radii, n, and denoting the " heaviness " of the substance by y^ we 
have y 7111"^ Ax as the weight of any disk, and ymc'Ax \_x + \Ax'\ 
as its moment about the axis Y. (The axis Y is perpendicular 
to the paper through 0.) The equation to the parabola being 
u^ = ^px, where p is the parameter, this moment may also 
be written 2 y 7rpx/ix[x -^- ^/Ix'], Sind hence the sum of all such 
moments for all the disks, or the total moment, may be expressed 
as 

M' = 2y7rp^''^l{x')']/]x + y7rp/Jx^^[{x)yx, . . (7) 

This sum does not apply to the paraboloid, but only to the 
stepped solid (or aggregation of disks with square edges), unless Ax 
is finally made zero. That is, the desired result (moment of actual 
paraboloid) will be obtained as the limit which M' approaches 
as Ax is made smaller and smaller. "When Ax is made zero, 

the first term of M' becomes ^ynp / {x^)dx ; (i.e., (p'{x) — x^) ; 

while as to the second, although the limit of 2^{x)Ax is 

I {x)dx, and hence is not zero, yet the outside factor, Ax, is 

zero and hence the second term vanishes. Therefore the total 

moment of the paraboloid \& M — ^ynp) I ''[x'']dx; and in this 



CENTKE OF GRAVITY. 



11 



we note that the (f)\x) of the general form of eq. (6) is a?^, and 
that the x anti-derivative of x^ is ^x^ -\- const. ; and hence, finally, 

3 



M = ^yniJ {V^^ {^x' + Oj = ^ynp \\x: -^ 



iyTtjoXn 



This quantity divided by the total weight (= y^tpx^') of the solid 
will give the distance of its centre of gravity from the vertex, or 



origin, ; 



I.e., x 



8^n 



Obelisk. 



As to notation, it is customary to anticipate the fact that the 

desired result justifies the use of the notation / "" \^(p\x)^dx, and 

to employ at once dx for ^x in making out the form of one term 
of the series, dx is then called an infinitesimal, which simply 
means that Ax is finally to become zero; in other words, that the 
result sought is the limit which the sum of the series approaches 
as Ax diminishes. 

14. Position of Centre of Gravity of Various Geometrical Forms. 
(Homogeneous, etc.; the plane figures representing thm plates 
of uniform thickness.) 

Ohelisk. Fig. 6 shows a homogeneous obelisk, or solid 
bounded by six plane faces, of which two are rectangular (hori- 
zontal in this figure) with cor- 
responding edges parallel (and 
hence these rectangular faces 
are parallel, and may be con- 
sidered as bases). Required 
the distance, s, of the centre 
of gravity, C\ of the obelisk, 
from the base EHGF. See 
Fig. 6 for notation. A is 
the perpendicular distance be- 
tween the bases. 

By passing a plane through the edge AD \\ to face BCGF\ a 
plane through AB \\ to JDCGH; and noting their intersections 
(dotted lines) with the faces of the obelisk, we subdivide it into 
the following geometric forms : 

A parallelojnped ABCD-JMGL, of volume V^ = Uh and 




Fig. 6. 



12 



NOTES AND EXAMPLES IN MECHANICS. 



having its centre of gravity at a distance z^ — \h above 

base IIF\ 
A triaiKjular pris7n AD-IJLH, of volume = T^^ = 2^(^i — ^^^ 

and for whose centre of gravity z^^ — ^h ; 
Another tnangxdcir prism AB-KFMJ^ of volume = F3 == 

\UJ)^ — Z')A and for whose centre of gravity z^ = \li ; and 

finally, 
K pyramid A-EKJl^ whose volume is Y^ = \^(f>x — ^)(A — 

and whose '' mean 2" is 2^ = ^Z'. 
Plence by eq. (3) of p. 19, M. of E., with 2^, etc., instead of 
x^ , etc., we have, after reduction, 

A 

%l _|_ ij^ _[_ j/^" ' 2 * 



- /),^, + Zhl + &,Z + l\ 

z — 



Triangular Plate. Fig. 7. Bisect .ij5 in M. Join 6>J/. 
Bisect OB in A^ and join AN, The intersection, 
(7, is the centre of gravity of the triangular plate or 
plane figure. [The centre of gravity of the mere 
perimeter of the triangle (slender wires, homo- 
geneous and of same sectional area) is the centre of 
the circle inscribed in a triangle formed by joining 
the middles of the sides.] 

Paraholio Plate. Fig. 8. AN being -\ to 

axis ON. X = fON. _ 

Upper Half of 2? receding Paraholic Plate, Fig. 9. x=^ON 
and y = ^AN. 






Fig. 9. 




Fig. 10. 



Semi-ellipse. Fig. 10. Semi-axes are a and l\ 



4^ 



Sector of a Spliei^e. AEBO, Fig. 11. Let li — the altitude 
of the zone, or cap, of the sector ; i.e., let h = r ~ 0K\ then 
0(7 = |r — |A ; C being the centre of gravity. 



SIMPSON'S EULE. 



13 



Segment of a Sphere. AK, Fig. 12. Let the altitude, AK, 



of the segment be A, and (7 the centre of gravity 

3 (2r-Ay 



then 



OG^ 



2>r-h 






Fig, 12. 

Any Pyramid {or Cone). Fig. 13. 




Fig. 14. 



Fig. 13. 

Join the vertex with 

Z>, the centre of gravity of the base MN. From D lay off 

Cl) = \D0. C is the centre of 

gravity of the solid. 

Zone on Surface of Sphere. Fig. 

11. (Thin shell, homogeneous and 

of uniform thickness.) The centre 

of g^ravity lies at the middle of the 

altitude, A, in the axis of symmetry. The small circles of the 

sphere, CB and AB. lie in parallel planes. 

15. Simpson's Rule (Fig. 15).— If ABCDEFG is a smooth 

curve and ordinates be 
drawn from its extremities 
A and G to the axis X, an 
approximation to the value 
of the area so enclosed, 
A..D..G..N..G ..A, 
between the curve and the 
N axis X, is obtained by 
.>j gi^ajjson'^s Rule^ now to be 
demonstrated. Divide the 

base OJSf mio an even number, n^ of equal parts, each = Ax (so 

that ON = n . Ax), and draw an ordinate from each point of 

division to the curve, the lengths of these ordinates being u^, 

u^ , etc. ; see figure. 




Fig. 15. 



14 NOTES AND EXAMPLES IN" MECHANICS. 

Consider the strips of area so formed in consecutive pairs ; 
for example, CDEE" C" is the second pair in this figure (count- 
ing from left to right). Conceive a parabola, with its axis vertical, 
to be passed through the points 6*, D^ and E. It will coincide 
with the real curve between G and E much more closely than 
would the straight chords CD and DE\ and the segment CDE^ 
considered as the segment of this parabola, has an area equal to 
two thirds of that of the circumscribing parallelogram CC'EE. 
Hence, since the area of this pair of strips = trapezoid 
GEE" C" -\- jparalolic segment GDE^ we may put 

Area of pair} ^ , V.r , n . «/ ,r , n^~l 

./ strip, CE" \ = ^^"^ L«''» + "*> + «(^^ - *t«^ + -J) j ; 

which reduces to ... \Ax\%i^ -\- 4:U^ -}- t^J. 

Treating all the ^ pairs of strips in a similar manner, we have 

finally, after writing Jx = {x^ — os^) -^ n^ 
Whole area \ x^ —x. 



AG" {approx.) \ - 3n L- ' 4^^+2«,+4^.,+2^*.+ ...+^.„ 

The approximation is closer the more numerous the strips and 

the more accurate the measurement of the ordinates ic^, u^^u^, etc. 

If the subdivision on the axis X were "infinitely small," an 

exact value for the area would be expressed by the calculus form 

rx - Xn nx = Xn 

I udx. Hence for any integral of this form, / tidx^ if we 

%J X — Xq kJ X = Xq 

are only able to determine the particular values (?/„ , u^ , etc.) of 
the variable u corresponding respectively to the abscissse a?^ , 
x^-\- Ax ^ x^-\- ^Ax, etc. (where Ax = {x^ — x^ ~ n, ii being an 
even number), we can obtain an approximate value of the integral 
or summation by writing 
rxn 

X ^"^^ " ^'' 

+ 2(.^, + .., + ... + .._,) + ^^J- 
As to the meaning of n^ note that the first ordinate on the 
left is not u^^ but ^c^ ; also that while there are 7% strips, the num- 
ber of points of division h n-\- 1, counting the extremities O 
and iT. 



udoe = \ - [u, + 4:{ti^ + ^^3 + . . . + '^« - 1) 




FOEM FOR ANALYTICAL CONDITIONS OF EQUILIBRIUM. 15 



CHAPTEE II. 

Principles and Problems involving Non-concurrent Forces 

IN A Plane. 

16. Most Convenient Form for Analytical Conditions of Equi- 
librium (Fig. 16). — Let Pj, P2, Pj, etc., constitute a system of 
non-concurrent forces in a plane acting 
on a rigid body and in equilibrium. Of 
the actual system, P^ and P^ are the 
only forces shown in the figure. As- 
suming a convenient origin, 0^ introduce 
into the system two opposite and equal 
forces, P/ and P/^, both acting at and 
equal and parallel to P^ . Evidently the 
presence of these two forces does not 
destroy the equilibrium of the original 
system. Similarly, introduce at the mutually annulling forces 
P^ and P^' bearing the same relation to P^ (parallelism and 
equality) that P/ and P/' do to P, ; and so on for each of the 
remaining forces of the system. Drop a perpendicular from 
on each of the forces of the original system, the lengths of these 
perpendiculars being a^ ^ a^^ a^^ etc. {a^ and a^ are shown in the 
figure). We now note that for each force P of the original sys- 
tem we have in the new^ system a single force at 0, equal and 
parallel to P and similarly directed, and also a couple, of moment 
Pa. For example, the force P^ of the original system is now 
replaced by the force P/^ parallel and equal to Pg and similarly 
directed, but acting at the j)oint ; and by the couple formed of 
the two forces P^ and P/, the arm of this couple being a^. It 
follows, therefore, that the new system consists of a set of forces 
(P/^ P/^, P^"^ etc.), all meeting at {and hence forming a 
concurrent system^ in a plane ^), and a set of couples, of moments 
Pi<^, , P^^a , etc. Since no single force can balance a couple 
(§ 29, M. of E.) or set of couples, the forces of the concurrent 
system at must be in equilibrium among themselves ; i.e. , ^and 
Y being any two directions at right angles, we must have 2X 

* And therefore, (unless balanced,) equivalent to a single force or resultant; 
see M. of E., p. 8. 



16 NOTES AND EXAMPLES IN MECHANICS. 

and 2Y separately equal to zero for the concurrent system at O; 
and the sec of couples must be in equilibrium among themselves, 
whence it follows that the onoment of their resultant couple must 
equal zero. Since the couples are in the same plane, the moment 
of their resultant couple is the algebraic sum (see § 34, M. of E.), 
i.e., P,a, + P,a^ + . . . , or :2{Pa\ = 0. 

Por the eqidlihriuin, therefore, of a system of non-concurrent 
forces in a plane, we must have not only ^X and ^ y = 0, hut 
also 2{Pa) = 0. That is, in practical language (the body being 
originally at rest), the forces of the system so ueutralize each 
other that they not only do not tend to move the body sideways 
or vertically, hut also do not produce rotation. 

In the practical application of these conditions of equilibrium 
in solving problems it is not necessary to introduce the pairs of 
equal and opposite forces at (the conception of which is needed 
only for purposes of proof), since the sum of the X-components 
(or y-components) of the actual forces of the system is equal to 
that of the X-components of the auxiliary forces introduced' at 
; while to form the moment-sum of the auxiliary couples, we 
have only to multiply each force of the actual system by the per- 
pendicular distance of its line of action from the origin, whose 
position is taken at convenience. 

The student should now read the latter half of p. 33, M. of E. 

17. The Rigid Bodies Dealt with at Present. — Each rigid body 
now to be considered is one whose dimensions perpendicular to 
the paper are supposed to be very small, and therefore may be 
considered to lie in the plane of the paper. An actual structure 
is made up of such pieces, or memhers, which are provided with 
forked joints, or duplicated in such a way that the above supposition 
(each piece lying in the plane of the paper) is practically justified. 

All surfaces of contact between any two contiguous pieces are 
supposed perpendicular to the paper, and friction between two 
such parts of a structure is disregarded ; i.e., the pressure hetween 
two contiguous pieces (or ^' members") is in the plane of the paper 
and normal to the surfaces of contact / for it is a matter of com- 
mon experience that pressure can be exerted at the sm,ooth sur- 
faces of contact of two bodies only in a direction 7iormal to those 
surfaces. 



CLASSIFICATION OF EIGID BODIES. 17 

In problems where the weights of one or more bodies con- 
nected with the structure are considered, the plane of the struc- 
ture will be vertical, and then (considering what has already been 
postulated) the system of forces acting on each member, or piece 
of the structure, is a system of forces in a ])lane (a " uniplanar" 
system of forces). 

18. Contact Forces or Pressures. — If one of the two bodies in 
contact is rounded at the point of contact, while the other is quite 
flat at that point, the action-line of their mutual pressure neces- 
sarily lies in a perpendicular, or normal, to the latter flat 
surface, and passes through the point of contact. Hence, the 
shapes of the bodies being known, this action-line becomes known 
on inspection. Let this be called '^ flat-co7itact pressicre^ (]^.B. 
As a better definition of flat-contact pressure, we might describe 
it as a pressure occurring in such a way that its action-line can- 
not he materially changed hy any slight motion of one piece rela- 
tively to the other ditring the small alteration of form and posi- 
tion which actually takes place when a load is gradually placed 
on the structure.) 

But if the mode of connection of the two bodies is a j^m- 
connection, that is, if one body carries a round pin or bolt fitting 
(somewhat loosely) in a corresponding ring forming part of the 
other body, we are unable to say in advance just where, on the 
inner circumference of the ring, the contact (and accompanying 
pressure) is going to be. Wherever the point is, all that we can 
immediately say as to the action-line of the force is that it passes 
through the centre of the circle, its direction (if determinate at all) 
being found from a consideration of the other forces acting on 
the piece in question. This will be illustrated later. 

Such a pressure may be called a hinge-pressure, 

19. Classification of Rigid Bodies under Uniplanar Systems of 
Forces. — As conducive to clearness in subsequent matter, the 
rigid bodies composing a structure will be named according to 
the number of forces acting on each ; and these forces consist of 
gravity actions (i.e., weights) and of the pressures exerted by 
neighboring pieces on the piece in question. The resultant gravity 
action on a single piece, or member of the structure, is a single 
force, called its weight, acting vertically downward through its 



18 



NOTES AND EXAMPLES IN MECHANICS. 




centre of gravity. (Of course, the action of gravity is distributed 
over all the particles of a body, but the above-mentioned single 
force is the full equivalent of these distributed forces as far as 
the equilihrium of the piece is concerned; though not such as re- 
gards the straining action on the piece. With these straining 
actions we cannot deal here ; they will be treated later in the 
proper place. In many cases the whole weight of a piece is so 
small in comparison with any of the other forces of the system 
acting on that piece that no appreciable error is made in regard- 
ing it as without weight. Notice is always given in such cases.) 
20. "Two-force Pieces" and their Treatment. — A "two-force" 
piece being a piece on which only two forces act, if the weight 
of the piece is considered there is but one other 
force. For example. Fig. 17, a body of weight 
G hangs by a stem and ring which form a rigid 
part of it, on a pin projecting from a fixed sup- 
port. 

Since, evidently, the equilibrium of a two- 
force piece requires that the two forces shall he 
equal and opposite and act in the same line, the piece A . . B 
will not be in equilibrium unless the centre of gravity c lies in a 
vertical line drawn through the centre of the circular section of 
the pin at A. Here we have an instance of the final full deter- 
mination (by the necessity of its being vertical) of the action- 
line of a hinge-pressure, concerning which we know in advance, 
only that it passes through the centre of the circle at A. We 
find, therefore, that the pressure of the pin at 
A against the ring of the rigid body A . , B, 
hanging at rest, must have a direction verti- 
cally upward, and an amount, F, numerically 
equal to G, while its action-line, c , . d, passes 
vertically through the centre of the hinge-circle. 
In Fig. 18 is another two-force piece, in 
which, for equilibrium, the same result is 
reached, but, the stem being curved, the straining action in it 
is of a bending nature ; whereas, in Fig. 17 it is a simple tension, 
or stretching action. 

The case of a two force piece whose weight is neglected is 




Fig. 18. 



THEEE-rORCE PIECES. 



19 



Bell Crank. 




Fig. 19. 



instanced in Fig. 19, where the two-force piece ^ . . ^ is sub- 
jected to tension through the 
action of a suspended weight, Gj 
and a bell-crank lever, BCD, 
The hiDge-pressure at A, from 
the support S against the piece 
A . , B, must pass through the 
centre of the circle at A ; while 
also the hinge-pressure at B, of 
the bell-crank against the piece 
A , . B, must pass through the 
centre of the hinge-circle at B. 
But these two hinge-pressures are the only forces acting on the 
piece A . . B, and for the equilibrium of a two-force piece 
must be equal, opposite, and coincident as to action-line ; that 
is, these pressures must both act in the line joining the centres 
of the two circles. A'B^ shows this piece represented as a free 
body, with the equal and directly opposite forces P' and P^ act- 
ing at the ends. As to the value, or amount, of this force P\ it 
cannot be found until the case of the bell-crank has been treated, 
depending, as it does, not only upon the design of the bell-crank 
and the amount of the load G, but also upon the position of the 
piece AB itself. 

21. Three-force Pieces. — If a rigid body is at rest (i.e., in 
equilibrium) under the actions of three forces, it is evident that 
these forces must have action-lines intersecting in a common 
point, and that each force must be the equal and opposite of the 
diagonal of the parallelogram formed on the other two as sides 
(laid off to scale) ; for any one of them imist he the anti-re sidtant 
of the other two. (See § 15, M. of E.) [In the particular case 
where the forces are parallel the intersection-point is at infinity 
and the value of any one of the forces is numerically equal to the 
sum of the other two (algebraic sum).] 

22. Example of a Three-force Piece. — The bell- crank BCD of 
Fig. 19 furnishes an instance. This body is subjected to the 
three hinge-pressures at B^ C, and 7>, respectively. That at D 
is a vertical downward pressure G\ equal to the weight G of the 



20 NOTES AND EXAMPLES IN MECHANICS. 

two-force piece DE. The action-line of the hinge-pressure at B 
has been found by the previous consideration of the two-force 
piece AB, and \^ a , .h. The hinge-pressure P'\ at C^ passes 
through the centre of the corresponding circle, but its action-line 
is as jet unknown. The problem, then, stands thus : Of the 
three forces, G\ P\ and P'\ under whose action the bell-crank 
is in equilibrium, G' is known in amount and line of action, P' 
is known as to action-line but not in amount, while as to P" we 
know neither its amount nor its direction but simply one point, 
C, of its action-line. However, since the three action-lines must 
meet in a common point, we need only note the intersection, 6>, 
of the known action-lines of P'- and G\ and join o with G 
(centre), in order to determine C , . o, the action-line of P^\ 
'Next, as to finding the amounts of both P' and P'^, consider that 
P^' is the anti-resultant of P^ and G\ and that therefore the 
(ideal) resultant of P^ and G' must act along c> . . G; hence 
lay off . . m by scale to represent G^ and through m draw a 
line II to <9 . . ^ intersecting o . , G in some point r, and draw 
7" . . ^ II to 6^^, to determine k on the line o . . B. Then o . . n, 
laid off along G . . Oy and = 6> . . r, but in the opposite direction 
from (9, gives the amount and direction of P" . For c> . . /• is 
the resultant of P' and G' ^ and o . .n\% its equal and opposite. 

Of course P' and P" must be measured by the same force- 
scale that was used in laying off o . . m = G' . 

We can see by inspection of the figure that if the position of 
the link, or two-force piece, AB^ were changed in such a manner 
that, while the line A . . B continues to pass through o, the pin- 
joint, or hinge, B is caused to approach nearer and nearer to (7, 
the forces P" (always equal to the ideal o . . ?') and P^ both 
increase without limit; for the point amoves out f rom 6>, m 
being fixed (i.e., the load G remains invariable) until, when B is 
infinitesimally near to (7, m . . r is || to o , . G and r is at 
infinity. 

The method just pursued is a graphic one ; analytically, we 
would proceed thus : since the system of forces G\ P\ and P^' 
is balanced, i.e., in equilibrium, the algebraic sum of their 
moments about any point in the plane must vanish, i.e., = 0. 



THREE-FORCE PIECES — ANALYTIC TREATMENT. 21 

(See § 16.) Take an origin, or centre of moments, at C and 
denote by d and c the lengths of the perpendiculars let fall from 
C upon the action-lines of G' and P\ respectively. (These 
lengths may be obtained trigone metrically from given distances 
and angles, but are most easily, and with sufficient j^recision, 
scaled off from an accurate drawing.) With G as origin the 
lever-arm of the force P" is zero and hence the moment of this 
force is zero ; consequently this force does not enter the moment- 
equation, which therefore will contain but one unknown quantity, 
viz., the amount of the force P' . 

The resulting moment-equation (following the routine recom- 
mended at the foot of page 33, M. of E.) is 

P\c- 6^'.6Z + P"x = 0..(1); whence P'^-G' ..,{1) 

c 

becomes known ; and since P" is the anti-resultant of P' and 
G', we have also, a being the angle between the action-lines of 
the latter forces. 




I 

I 

-»!c 



P" = VP" + G" + "iP'G' cos Of (2) 

(See p. 7, M. of E.) 

23. Other Examples of Three-force Pieces. — The ordinary 
straight lever, with flat-contact supports, is shown in Fig. 20. 
Since the pressures (or reactions) of the 
supports against the lever must be "| 

to the axis of the latter, and hence ^ i i;;:;;:;- j/:.>,.;:::^^:/.::;:/;/.:.:^v;-v:.-.:i;:| 
parallel, in this case the action-line of 
the third force P' must be made "j to 
the lever. Otherwise equilibrium could ^^^o- 20. 

not he maintained, for the point of intersection of the three force- 
lines is fixed by the intersection of the flat-contact pressures 
P" and Q ; at infinity in this instance. 

Given P' , we determine Q and P" by considering the lever 
as a free body under a system of three forces in equilibrium (in 
a plane), taking a moment-centre at B so as to exclude the 
unknown force P" from the equation ; and obtain first, as a 
moment equation, 

Qh-P'a=^% . . or, Q = jP'', 

b 



22 NOTES AT^D EXAMPLES IN MECHANICS. 

and then, bj summing all the components of the three forces in a 
direction at right angles to the lever, 

+ P'' - Q -F' = 0', whence P" ^ Q ^ P' -^ 

and thus the two unknown forces have been found in amount 
(and are already known in position). 

Obviously, if Q is given, being, e.g., the weight of a body to 
be sustained, we compute in a similar manner the necessary force 
P' to be applied at C and the resulting pressure, P" =^ P' -\- Q^ 
at the support or fulcrum B. 

It is also evident that the smaller the distance h is made in 
comparison with a, the greater the pressures P" and Q^ for a 
given P' ; in fact, as h approaches zero, P'' and Q increase with- 
out limit. For a given Q and diminishing value of the ratio 
h ; a, the necessary force P' decreases toward zero. (K.B. In 
these cases the weight of the lever itself is neglected.) 

As showing how the possibihty of equilibrium may be de- 
pendent in some cases on the design and position of the support- 
ing surfaces, let us consider the curved lever 
in Fig. 21, where the supporting surfaces A 
and B are capable of furnishing only flat- 
contact pressures or reactions, whose direc- 
tions, A . . and B . . O, are fixed, being 
normal to the respective smooth and flat 
surfaces of contact. (E".B. Smooth sur- 
faces are postulated in all the present prob- 
lems ; rough surfaces will be considered 
later.) 

The intersection, O, of their action- lines is therefore fixed, 
and if a force P' is to be applied at a given point C to induce 
pressures at A and B, its line of action must be taken along 
C . . 0'\ otherwise the lever will begin to move out of its pres- 
ent position (weight of lever neglected). Given, then, the force 
P' along C > . 0,WQ determine P" and Q for equilibrium, in the 
same manner as before shown, by filling out the parallelogram 
, m . r , hm Fig. 21, precisely as was done in Fig. 19 (except 
that the P' of this problem corresponds to the G' of Fig. 19). 




CUEVED LEVER. REDUNDANT SUPPORT. 



23 




However^ if we assume any direction at pleasure for the 
action-line of P' through the point G^ and wish to secure equilib- 
rium, we have only to change the mode of 
support at A or at ^ (say B\ into a pin- 
joint or hinge-support ; for the direction of 
a hinge-pressure is not determined solely 
by the nature of this mode of support ; the 
only restriction upon it known in advance is 
that it must pass through the centre of the 
hinge-circle, its direction being determined 
by other relations. This change being made, 
we have Fig. 22, in which P' is given, or ^m. 22. 

assumed, both in amount and position. At A we have an un- 
known flat-contact pressure ^ in a known action-line A . . O, 
The intersection O of the two action-lines A . . and C . , must 
be a point in the action-line of jP'^, the unknown hinge-pressure ; 
i.e., P . . is the action-line of the latter, while its amount, as 
well as that of Q, is found by a construction like that in Fig. 19, 
which need not be explained again. The results are that 

P'' =^ . . Jc, and Q = . .n (the equal and opposite of (9 . . r\ 

these lines, or rather lengths, representing forces on the same 
scale as that on which O . . m represents the first given force P\ 
24. Redundant Support.— If the two supports, A and P, of 
the lever are both hinge- joints, as shown in Fig. 23, the body or 
lever APG is redundantly supported ; for 
now the hinge-pressures at A and P are 
indeterminate, from simple Statics alone, 
but depend on the form and elasticity of the 
lever itself and upon the degree of loose- 
ness of fitting of the hinge-rings around the 
pins of the supports A and P, and upon 
any slight elastic yielding of the latter ; as 
well as upon the amount and position of 
the force P'. 
In fact, if the body is elastic, and we have to " spring" it to 




Fia. 23. 



24 



NOTES AND EXAMPLES IN MECHANICS. 



cause the rings to fit over the pins, pressures are produced at the 
supports 'before the application of any force at C. 

From simple Statics, then, all that we can claim is that the 
action-lines of the hinge-pressures P" and Q must intersect in 
some point situated on the given action-line of the given force 
P' ; but have no means of fixing the position of this point 0. 
Problems of this nature, therefore, cannot be treated until the 
theory of Elasticity is presented, and then, as will be seen, only 
in comparatively simple cases. In attempting an analytical treat- 
ment in this case we should find that it presents y6>i^r unknown 
quantities ; whereas from simple Statics only three equations 
(independent equations) can be obtained for the equilibrium of a 
system of forces in a plane ; hence the indetermination. 

25. Four-force Pieces. — If the rigid body is in equilibrium 
under a system of four forces one of which is given both in 
amount and position, while the action-lines of all the other three 
are known, the amounts of those three can be determined. 

A simple graphic method for solving this case is based on 
the obvious principle that if four forces are in equilibrium the 
(ideal) resultant of any two must be equal and opposite to the 
resultant of the other two and have the same action-line. 

26. The Simple Crane. — A convenient example of a four-force 
piece is presented by the simple kind of crane, APC, in Fig. 24, 

consisting of a single rigid body, of 
curved form. Its lower extremity rests 
in a shallow socket, while at B the edge 
of the (wharf) floor furnishes lateral 
support. We neglect the weight of the 
crane, and assume that no pressures are 
induced at A and B unless the crane 
bears a load ; i.e., that the parts are 
loosely-fitting at A and B. Placing 
now a known load at 0, viz., P, , we note that in preventing the 
overturning of the crane the right-hand edge of the floor at B 
reacts against the crane with some horizontal pressure P„ 
(horizontal, since the surface of contact is vertical), while at A 
there are two surfaces under pressure, one of which is horizontal^ 




Fig. 24. 



SIMPLE CRANE — GEAPHICALLY AND ANALYTICALLY. 2o 

while the other is the left-hand vertical side of the socket ; there- 
fore at A we have the vertical and horizontal reactions, P, and 
P^ , both unknown in amount. 

Now pair off the four forces at convenience ; for example, as 
in our figure, pair off P, with I^^ , noting the intersection, o, of 
their actionJines ; P^ and P^ constitute the other pair and inter- 
sect at o\ 

Since {? is a point in the action-line of the resultant of P, and 
P^ , while o' is a point in that of the resultant of P, and P^ ; 
and since these two resultants must have a common action-line 
for equilibrium, that common action-line must be c> . . o\ Pro- 
ceed tlierefore as follows : Prolong O . . o and make o . . 7n equal 
to I^^ by any convenient scale. We know that the diagonal 
of the parallelogram formed on P^ and P^ must lie on the line 
o . . o' (prolonged, here). This parallelogram is found by making 
m . . r II to P^ to determine rono..o''^ then drawing r . . ^ || to 
til . . to intersect ^ . . c> in some point Ic. JN^ow prolong o , . o' 
beyond o\ making o\ , r' equal to o , . r. Then o' . . r' is the 
resultant of the unknown P^ and P^^ , and by drawing the proper 
parallels, as shown, we resolve d . , r' in the directions of those 
forces and thus determine o' . . ni' = P^^ and o' , . ^' = i^^ ; 
o . .h^ already found, is equal to P^ ; all, of course, on the as- 
sumed scale of force (which scale is entirely independent of the 
scale for distances used in laying out the dimensions of the crane 
on the paper). 

Evidently, in this particular problem, P^ = P3 , and P^ = P^. 
(KB. It will be noted that by this method the directions of 
pointing of the unknown forces are found, as well as their magni- 
tudes.) It is also easily noted, on inspection, that if the distance 
B . , A is made shorter and shorter and diminishes towards 
zero, Pj remaining the same in amount and position, the forces 
P2 and P4 increase without limit (i.e., become ififaiite for 
B ..A = 0). 

27. Simple Crane Treated Analytically. Single Rigid Body. — 
Take ^ as a centre of moments (for by so doing we exclude the 
two unknown quantities P, and P^ from the moment-equation, 
since then their lever-arms are both zero : and thus obtain an 



26 NOTES AND EXAMPLES IN MECHANICS. 

equation containing only one unknown quantity). The lever- 
arm of P\ is Ao\ and that of P^ , Ao, Hence 

. . JT 

+ P, . ^^' - P, . ^0 + + = ; whence P, = = . P,. . (1) 

Ao' 

The balancing of horizontal components gives 

+ P,-P, = 0, whence P^^P,', i.e, P, = ^f^. P,. . (2) 

Ao' 

Also, from the balancing of vertical components, 

P.-P. = 0;i.e.,i'3=P. (3) 

28. Multi-force Pieces. — Since we can at most determine the 
amounts of only three unknown forces (action-lines given) in 
a uniplanar system in equilibrium; and since to determine any 
forces at all there must be one force given in all of its elements 
(i.e., amount and position) ; it follows that when the system con- 
sists of more than four forces, all but three of them must be given 
in amount and position, and also the action-lines of the unknown 
three must be given, if the problem is to be a determinate one. 
In such a case it is a simple matter to combine the known forces 
into a single resultant by successive applications of the parallelo- 
gram of forces, and thus reduce the problem to that of a four- 
force piece, treating it then as in the last figure (Fig. 24), the 
resultant of all the known forces playing the part of P, in that 
figure. (N.B. To find graphically the resultant of two jjarallel 
forces we can use a construction like that in Fig. 10 or 11 of pp. 
13 and 14, M. of E.) 

29. Compound Crane on Platform-car.- — As an example of the 
itility of the foregoing principles, let us apply them to the case of 

the several rjgid bodies constituting the composite (or built-up) 
crane of Fig. 25. 

Here we have an assemblage of seven rigid bodies, forming 
a rigid structure at rest ; viz., the jih^ ABC \ the tie-rody DB ; 
the masty TJADE\ the tie-rod^ FE\ the platform^ P'S; and the 
two wheel-pairs, M and P. (Each pair of wheels and its axle 
form together a single rigid body.) The track is level. 

For simplicity we shall consider the platform alone as having 
weight, viz., a force 6^, , applied in the centre of gravity, as 



COMPOUND CRANE ON CAR — WHEEL-PRESSURES. 



27 



shown ; while the extremity C of the jib is to carry a load G, 
We have given^ therefore, the two forces G^ and G^ and all 
angles and distances concerned, and are required to find the 
pressures induced at the various points of contact between the 

Compound Crane. 




>rr.^s.til 'J?^^S?^^^ 



Fig. 25. 



parts of the structure, and under the wheels. (The final practical 
object is, of course, to give sufficient strength to the parts for 
their respectiv^e duties, a matter, however, which cannot be 
entered into here, belonging, as it does, to the topic of " Mechan- 
ics of Materials.") Only analytical methods will be used. 

29a. The Wheel-pressures. — Let us first consider the whole 
structure as a free body. The only forces external to this free 
body are the two gravity forces G and G^ , and the vertical up- 
ward pressures, Y^ and Y^^ , of the rails against the wheels ; these 
constitute a system of parallel forces in equilibrium and are all 
shown in Fig. 25. (I^ote. — The pressures between any two con- 
tiguous parts of our present free body do not need to appear in 
this system for the reason that, if introduced, they would form a 
balanced system among themselves and might therefore be re- 
moved without affecting equilibrium. For example^ at D the 
ring of the tie-rod presses horizontally and to the right against 
the pin of the mast ; and the pin, from the principle of action 
and reaction, presses the ring with an equal horizontal pressure 



28 NOTES AND EXAMPLES IN MECHANICS. 

toward the left ; but hoth the ring and the pin belong to the 
free body now under consideration in Fig. 25, and if one of 
these pressures is inserted in the system the other must be placed 
there with equal right, and the two then annul each other'' s influ- 
ence in all the equations of equilibrium. We therefore conclude, 
in general, that the mutual actions between the parts of a given 
free body in equilibrium may be omitted in applying the con- 
ditions of equilibrium^ 

With 6^ as a centre of moments, then, we have for the free 
body of Fig. 25 (see figure for notation) 

VnCin — G{a + m) — G,a, = ; .'. F^ = -^ — '—^ — :^ ; 

a^ 

while by summing vertical components 

F,+ 7.-6^-6^, = 0; hence Y;=G+G,-Y^. 

29b. Pressures at the Joints ; D, B, and A (Fig. 25). — By in- 
spection we see that the tie-rod DB is a two-force piece (its own 
weight neglected) ; that is, the hinge-pressures at D and B must 
have a common action-line, viz., J) . . B. DB is a straight two- 
force piece / or '' straight lin'k^'' as we shall hereafter call it ; and 
is subjected to a tensile action along its axis (tension, here ; as we 
note by inspection ; a straight link under compressive action is 
called a strut^ or compression-member). 

The Jib as a Free Body. Fig. 26. It is a three-force piece, 
I being acted on by the known vertical down- 

ward force G at 6^, by an unknown horizontal 
force T (directed toward the left) at B {T 
being the pull of the tie-rod), and an unknown 
hinge-pressure P at J., making an unknown 
angle a. with the horiozntal. 

Note. — Since the mast which acts on the 

jib at A is not a twoforce piece, we have no 

A' ^h means of knowing the position of the action- 

Fia. 26. I'j^g ^£ p ^Yoxa a mere inspection of the mast, 

as we did with the tie DB and the hinge-pressure at B. Of 

course, graphically, P passes through the point on where the 

action-lines of the other two forces intersect ; but as we are now 



TENSION IN THE TIE-ROD. 



29 



using analytical methods, we shall replace P^ which is unknown 
in amount and in position, by its (ideal) horizontal and vertical 
components, P^ and P^ (i.e., by two unknown forces in hnown 
action-lines). We thus have a four-force piece to deal with. 

If we take ^ as a centre of moments, the force T will be the 
only unknown quantity in the corresponding moment-equation, 
which is Th — Ga = ; whence we have, for the tension T in 
the tie-rod, T = G{a ~ h). (This T is the value of the hinge- 
pressure at P and also that at P, in Fig. 25.) 

Assuming an axis ^horizontal and JT vertical, we now have 

from :SX= 0, Pj^- T=0; or, Pj,= T; while from 

^ y = 0, Py — G — ; or, P^ = G; and we now 

, P , 

while tan a ^= -j^. 



easily find P itself, since P — VPu + Pv 



29c. Tension in the Tie-rod FE. As with DP, so with FP 
we note by inspection that it is a two-force 
piece, so that the hinge-pressure, T\ at E, 
Fig. 25, though unknown in amount as yet, 
must have P. .Pas action-line. The force 
T^ and the pressures (or supporting forces) 
iZ'and F exerted by the left-hand side and 
bottom, respectively, of the shallow socket 
U against the foot of the m.ast, are the three 
unknown forces in known action-lines acting 
on the free body shown in Fig. 27, consist- 
ing of the jib, mast, and tie-rod PP. The 
mutual actions between these three bodies are internal to the 
free body taken and are hence omitted (see note in § 29*^), the 
external system consisting of T\ H^ F, and G', all in known 
action-lines, G being the only force known in amount. By 
moments about the point U^ we have 




Tn - Ga = 0', whence T' = -• G. 



n 



(1) 



From ^X=0, ZT-r' cos «:'=::: 0, an^ r . R^-G cos a'. (2) 



From:^F = 0, V-G-T'sma' = 0\ .-. Y=^G+T sma'. (3) 



NOTES AND EXAMPLES IN MECHANICS. 




It will be noticed that we have not made use of the mast as a 
separate fiee body. This might have been done as a means of 
finding the three forces just determined, T\ J7, and Y\ since the 
hinge-pressures at D and A had already been de- 
duced ; but the process would have been more 
roundabout. 

However, as a reminder of the principle of 
action and reaction and of the definition of force, 
Fig. 28 is presented, showing the system of forces 
we should have to deal with in treating the mast 
as a free body ; and also Fig. 29, representing the 
car-platform and the two pairs of wheels as a single 
free body, with the external forces acting. The 
student will note that the R and Y in Fig. 28 are f^^- ^s- 

the equals and opposites, respectively, of those in Fig. 29. A 
similar statement may be made for the T' of those figures. 

Again, the P and T of Fig. 
28 are the equals and oppo- 
sites of the P and T of Fig. 
26 (action and reaction). The 
plane of the crane being sup- 
posed to be midway between the two wheels of each pair, the 
pressure Y^ is equally divided between the two wheels forming 
the pair on the left. Similarly at iV, with Y^ . 

30. Simple Roof and Bridge Trusses ; Ritter's Method of Sections. 
— A truss is an assemblage of straight pieces jointed together in 
one plane. If the joints consist of pins (one in each joint) 
inserted through holes or rings in the ends of the pieces (or 
*•' members") the truss is said to be '-'- pin-connected'''' ; while if the 
ends of the pieces meeting at each joint are rigidly riveted 
together (a favorite method in Europe) the truss is said to have 
riveted connection. 

In the first case, pin-connection^ each piece is free to turn 
about the pin, independently ef all other pieces, during the 
gradual, though shght, change of form which the truss undergoes 
in the gradual settling of a load upon it, and the stresses induced 
in the pieces are called " primary stresses" (whereas, with riveted 




Fig. 29. 



EITTEK'S method of sections — EXAMPLE OF TRUSS. 31 



joints, other, and additional, stresses, called " secondary stresses," 
are caused in the pieces, from the constraint exerted on each 
other by the members meeting at each joint). 

Confining ourselves to the consideration of pin-connected 
trusses, constructed so that each piece connects no more than two 
joints^ and loaded only at the joints (its own weight being con- 
sidered as concentrated at the various joints), we note that in 
such a case each member must be a straight two-force jpiece^ or 
straight linh (neglecting its own weight) ; i.e., it is subjected to a 
simple tension or compression (according as it is acting as a tie 
or a strut) along its axis. 

]S"oTE. — If such a straight link be conceived divided into two 
parts (for separate treatment) by any imaginary transverse plane 
or surface passing between the joints connected by that piece, 
the action or force exerted on one of these parts by the other is 
a pull (if tension) or thrust (if compression), applied at the section 
and directed along the axis or central line of the piece. For 
exanijylt^ in the truss of Fig. 30. pin-connected, and composed of 



a-^T 





Fig. 30. 



Fig. 31. 



" straight links," if we wish to consider free the portion on the 
left of the imaginary cutting surface A , . B^ the system of forces 
acting on the ideal body so obtained (see Fig. 31) consists of the 
abutment reaction Y^ , the loads G^ G\ and G" (at the joints 
<z, 5, and c\ and the three forces (pulls or thrusts) P, Q, and T, 
acting at the (cut) ends of the three pieces intersected by this 
surface A , . B. 

If the pier reactions F^ and F„ have been already determined 
by a consideration of the whole truss as a free body, F^ is now a 
known quantity, and we may go on to find the values of the 



32 NOTES AND EXAMPLES IN MECHANICS. 

three stresses (i.e., pulls or thrusts) P^ Q^ and T^ by methods 
already illustrated. It is evident from inspection that the force 
or stress in the lower horizontal piece is a pull (P) ; and that the 
stress {T) in the upper horizontal piece is a thrust ; but as to 
which way the arrow indicating the stress Q in the oblique 
member (" web-member") should be pointed (a matter not always 
to be decided on mere inspection) the detail of the analysis would 
show, provided numerical data were given throughout. Examples 
of the application of Emitter's method will be given later. 

This method is peculiarly well adapted to the treatment of 
pin-connected, straight-linked, trusses, since the stress in a straight 
link (its own weight being neglected) is always a simple tension 
or compression ; in other words, a pull or thrust directed along 
the axis of the piece. 

The three stresses, P, Q, and T, which were stated to be 
obtainable from the free body in Fig. 31, in the three straight 
links concerned, could also be determined from the consideration, 
as a free body, of the other jportion of the truss^ viz., that on the 
right of the cutting surface A , . B of Fig. 30 ; in fact, in the 
present case with greater simplicity^ since in 
y^i this new free body, shown in Fig. 32, the sys- 
tem of forces in equilibrium is much simpler, 
though there is, of course, the same number of 
JD unknown quantities, jP, Q^ and T; which, it is 
Lr to be carefully noted, are the equals and oppo- 
FiG. 32. sites, respectively, of the P, Q, and T of Fig. 31. 

31. Remark. — From the foregoing illustrations (dealing with 
compound quiescent structures bearing loads) we note that in the 
various free bodies (whose conception has been necessary for 
introducing the different unknown forces into balanced systems) 
the pressures or pulls of any two parts of the same free body 
against each other are omitted from the system ; and also (see last 
paragraph) that a portion of a two-force piece, situated at one 
end thereof, may be conceived removed, and the pull or thrust 
of that portion against the remaining portion inserted in the 
system, provided it is a straight two force piece, whenever it is 
desired to consider separately either part of a pin-connected truss, 




PROBLEM OF THE TWO LINKS. 



33 



conceived to he divided into two jparts hy a cutting plane or sur- 
face (see the last three figures). (The stresses in bodies that are 
not straight links will be considered later in the '* Mechanics of 
Materials:' pp. 195-514, M. of E.) 

32. Problem of the Two Links (Prob. 2 and Fig. 36 of p. 35, 
M. of E.). — Consider link AB free in Fig. 33. Since this is not 
a two-force piece, the action-lines 
of the hinge-pressures at the ex- 
tremities do not coincide, nor does 
either follow the axis of the piece. 
Hence these pressures are best 
represented by their horizontal 
and vertical components, as shown, 
so that the four unknown quanti- 
ties, Xo , J^o , X, , and y, , are to 
be determined. Similarly, considering the other link as free, we 
have Fig. 34, in which it is to be noted that the components of 
the pressure at the upper hinge are respectively equal and oppo- 
site to those {X^ and Y^) of Fig. 33 (action and reaction), so that 
there are only six unknown ''force-amounts" in the two figures, 
instead of eight, as might appear at first sight. Hence the prob- 
lem is determinate, since from each of these free bodies we obtain 
three independent equations; or six in all, as follows: Putting 
JS'X^' 0, ^r = 0, and '^{Pa) = (i.e., :S moments), for each 
body in turn, taking the centre of moments at the lower hinge in 
each case, we have 




Fig, 33, 



For 



^^^' "^^ ) and XJi + Yfi - G,a, - G.a, = 0. 



For Fig. 34 



and Yh' 



^.-0; Y, 
X,h' + Cuh 



0. 



The elimination, by which to find separately the six unknowns, 
Xo, Zq, X, , 1^1, X,, and Z,, is left to the student. (Treated 
graphically, this case would come under Class B of p. 458, 
M. of E.) 



34 



NOTES AND EXAMPLES IN MECHANICS. 




Fig. 35. 



33. Problem of Rod and Cord (Prob. 4 and Fig. 41 of p. 37 
of M. of E.). — Consider the rod free by 
cutting the cord and removing the pin 
of the hinge ; that is, besides the forces 
G^ and G^^ we must insert the unknown 
tension P along the axis of the cord, at an 
angle oc with the rod, and the horizontal 

and vertical components, X^ and Y^^ of the hinge-pressure, whose 
action-line is unknown, and thus have a complete svstem of forces 
in equilibrium. There are only three unknown quantities, F\ 
X„,andr„. 

From '^ hor. comps.=0, we have X^—P' cos o-^O ; . ' . . (1) 

" :^vert. " =0," '^ Y,-^P'^\vLa-G-G,=^\{4) 

^ (moms, about hinge) . . . P'c — G^a^— G^a^=^0. ... (3) 

Since eq. (3) contains only one unknown, P\ we have at once 

P'={^G,a,^G,a,)-^c', 

and knowing P\ we obtain X^ from (1), and Y^ from (2) ; and 

finally the hinge-pressure, P = VX^' -j- Y^% making an angle 

whose tan = Y^ -^ X^ with the horizontal. 

34. Problem of Simple Roof-truss (Prob. 5 and Fig. 40 of p. 
37, M. of E.).~The right-hand 
support is supposed to furnish all 
the horizontal resistance. Hence 
the system of forces acting on the 
whole truss, considered free, will 
be as shown in Fig. 36, in which 
there are three unknown reac- 
tions (or pressures, of the support- 
ing surfaces), Fo , Vn, and IT. H 
becomes known from 

:S'X=0, viz., 2Trsin«' 
By moments about point 0^ we have 

FJ -PJ.-P,,\l- IFJ + + + = 0, 
which can be solved for F^ , while from ^ moms, about B 

^YJ^^PIJ^W .1 cos a + TF(Z cos a: - Z^) + P, . iZ = 0, (3) 
from which F can be obtained. 




Fig. 36. 



/^=0. 



(1) 

(2) 



SIMPLE ROOr-TKUSS. 



35 



[Note. — HaviDg now made use of three independent equations, 
based on the laws of equilibrium of forces, and by their aid de- 
termined three quantities originally unknown, the student should 
not imagine that by putting 2 Y = 0, or by writing another 
moment summation about the point A (for example), lie thereby 
secures another independent equation, from this same free body, 
capable of determining a fourth unknown quantity. He would 
find that such an equation could be established by mere algebra, 
from the tirst three above, without further reference to the figure, 
and hence would be useless as regards determining any other 
unknown quantities. See top of p. 33, M. of E.] 

All loads or forces being considered to act at the joints, 
and no piece extending beyond a joint, we note that this roof- 
truss is composed entirely of straight two-force jpieces (each in 
simple tension or compression along its ^r - -•§ 

axis), so that portions of the truss may be \ 
considered free, isolated by the passing ot 
one or more cutting surfaces. For ex- 
ample, to find the stress in piece A and 
that in CO consider the free body in Fig. 
37, where S and T are the stresses required 
figure form a concurrent system, for which 




Fig. sr 

The forces in this 



-^7-- 



:2'X= Ogives xS COS £y+r cos /5-fTF" sin a=0, ... (4) 
and :^ Z = " S sin or-f T sin /?+ Y-P- TTcos a=0. . (5) 

Solve these for >S' and T. In a numerical case one or both of these 
I will come out negative, indicating compres- 

sion, not tension as assumed in the figure. 
To find the stresses m AC and AB, the 
free body shown in Fig. 38 may be taken. 
Here the forces form a non-concurrent sys- 
tem. Taking moments about B, we have 
7a- ?7. 1^ + + = 0; . (6) 
from which, T being already known, U can 
be obtained. For JR^ put 2Y =0^ and it will be the only un- 
known quai^tity ; or, put moments about 6^ = 0. 




Fia. 38. 



36 NOTES AND EXAMPLES IN MECHANICS, 

35. Problem. Rod and Tumbler (Fig. 39).— The tumbler is 
^ ^ , smooth-edged with vertical sides. The rod 

P has smooth sides and weighs G lbs. C is 
its centre of gravitj^ at a distance a from 
the end of rod. Given the distances a and 
d^ at what inclination a with the horizontal 
-^, should the rod be placed, in contact with 
I tumbler at two points as shown, that the 

hm//r/mmm/i/h/r//m/7/jr Position of the rod maj be stalle, i.e., that 
Fig- 39. -fche rod may remain in equilibrium ? G^ 

a, and d are known ; H, P, and a unknown ; H and P being 
the pressures of the tumbler against the rod at the two points of 
contact. H must be horizontal (why ?), and P "1 to side of rod 
and hence at angle a with the vertical. 
The rod being the free body, 

from :2X = we have H — P ^m a = 0', (1) 

'' :eY = " " - 6^ -f P cos a: = ; . , . . . (2) 

" 2 (moms, about O) we have Pd sec a — Ga cos a=^0. . (3) 

[N'ow sec a: = 1 -^ cos ^, and, from (2), G= P cos ol. Hence 

(3) becomes 



d 

— Pa cos'^ (X = ; . *. cos a 



3 Id 



cos a 

a being now known, P and H are easily found from (2) and 
(1). (A brief mode of finding a alone is based on the fact that 
the three action-lines concerned must meet in a common point m. 
If, therefore, a figure be drawn in which the action-line of P in- 
tersects that of ^ in a point m' not coincident with m, that of H 
and G^ we have only to form a trigonometrical expression for the 
distance mwi', involving a, d, and «, write it = 0, and solve for 
cos a. or sec <^.) 

36. Problem. Pole and Tie (Fig. 40).— Given the load P, the 
weight G of the pole, and all the distances and angles marked in 
the figure, it is required to find the tension P' induced in the 
chain, whose weight is neglected, and which thus serves as a 
straight tie. The pole is hinged at 0. 

The pole may be considered free, as already shown in the 



POLE AND TIE. 



37 




Fig. 40. 



figure, bj inserting tlie pull P' exerted on it at ^ in a known 
action-line, h . . l^ by the chain, and the 
horizontal and vertical components, X^ 
and Y^ , of the hinge-pressure at 0. The 
action-line of this hinge-pressure makes 
an unknown . angle with the horizontal, 
but must pass through the centre of 
the hinge at 0. There are three un- 
knowns in the system, viz., X^, Y^^ and 
P'. 

From '2X = we have 

P' cos ^ - X„ = ; . . (1) 
" JST^O " " Y-P-G— P' sin a = 0;. . (2) 
" 2 (moras, about 0) = 0,Pa+ Gh -P'a' = 0. . . (3) 
The value of P' is easily found from (3), then that of X^ from 
(1), and that of Y^ from (2). Hence the amount of the 
hinge-pressure, P^ = \^ X^ + ^o^ becomes known, and the tan- 
gent, = Zf, -^ Xo , of the angle between its action-line {On) and 
the horizontal. 

GrapMc Solution. The action-line of the resultant, P^ 
= P -f- 6r, of the known parallel forces P and G may easily be 
determined by a construction like that of Fig. 10, M. of E. ; or 
by applying the principle of the foot-note, p. 14, M. of E. Since 
this action-line intersects that of the force P' in some point 
n, the hinge-pressure at must act along the line , , n and 
must be the equal and opposite of the resultant of P^ and P. 
37. Problem. Three Cylinders in Box (Fig. 41). — Three solid 
homogeneous circular cylinders, of equal 
weight, = G, and of the same dimen- 
sions, rest in a box, as shown, of hori- 
zontal bottom and vertical sides. Th^ 
two lower cylinders barely touch each 
other ; i.e., there is no pressure between 
them, as the box is an easy fit. The 
W///////////II////II/M///I/IM centres of the three cyliDders form the 
Fig. 41. vcrticcs of au equilateral triangle. It is 

required to find the pressures at all points of contact {points, in 



Three Ci/linders. 




88 



NOTES AND EXAMPLES IN MECHANICS. 




Fig. 42. 



Fig. 43. 



this end mew / really, lines of contact) between the cylinders and 
the box ; also between the cylinders themselves. All surfaces 
smooth. 

Fig. 42 shows the upper cylinder as a free body, there being 
three forces acting on it, viz., G^ the action 
of the earth, or gravity, and the two press- 
ures, IP' and P" ^ from the cylinders be- 
neath. From symmetry, P' and P" must 
be equal. From ^ (vert, comps.) = 0, 
^P' cos 30° - 6^ = ; 
i.e., P' (or P") = G-^V^. 
Taking now the lower right-hand cylinder free in Fig. 43, we 
find it under the action of its weight G ; of the pressure P' (the 
equal and opposite of the P' in Fig. 42) now known ; of the un- 
known horizontal pressure or reaction P'" from the side of the 
box ; and of the vertical pressure P^ from the bottom, also 
unknown. (Concurrent system, with two unknown quantities.) 
There is no pressure at A, (See above.) From 2 (hor. 
comps.) = 0, 

P' sin 30° - P"' == ; .-. P'" = G{ VS -^ 6). 
From 2 (vert, comps.) = 0,P,-G -P' cos 30°= ; .-. P„= %G. 
It thus appears that the sum of the two pressures on the 
bottom of the box is ZG^ i.e., is equal to the combined weight 
of the three cylinders ; if the sides of the box were not vertical, 
however, this would not be true, necessarily. 

38. Problem of the Door. — Fig. 44 shows an ordinary door as 
a free body. Support is provided by two verti- 
cal hinge-pins, of smooth surfaces, whose dis- 
tance apart is such that the lower one alone 
receives vertical pressure from the door (i.e., 
furnishes a vertical supporting force, Y^ , at its 
horizontal upper face). The upper hinge-pin pro- 
vides only lateral support, as seen in the horizon- 
tal reaction H which prevents the door from fall- 
ing away to the right, from that hinge, Simi- 
FiG. 44. larly, the right-hand vertical edge of the lower 




DOOR — WEDGE AND BLOCK. 



39 



hinge-pin, hy its reaction 11^ , offers lateral support at that point. 
Given the weight G of the door (considered as concentrated in 
its centre of gravity) and the distances a and A, it is required to 
determine the three pressures, JI, H^ , and Y^ . 
From ^2 (vert, comps.) = we have F^ — (x = ; i.e., Y^ = G. 



From 2 (moms, about 0), Hh — Ga = ; i.e., 



n^la. 



From 2 (hor. comps.) = 0, M — H^ ^= ; i.e., H^ 



h 



G. 



Since Y^ is parallel and numerically equal to G, and H to 
H^, the system of forces acting on the door is seen to consist of 
two couples of eqnal moments of opposite sign, thus balancing 
each other. The smaller the distance h the greater the value of 
H (and of its equal, H^^ if G and a remain unchanged. 

(Unless the fitting of the parts is very accurate, only one 
hinge of a door receives vertical pressure — i.e., carries the weight, 
in practical language.) If more than one receives vertical press- 
ure, the share carried by each depends on the accuracy of fitting 
and on the slight straining or change of form of the parts under 
the forces acting.) 

39. Problem of the Wedge and Block (Fig. 45). — The shaded 
parts represent a smooth horizontal table 
or bed-plate eh^ and a flat and smooth ver- 
tical guide md, both immovable. W is 
the wedge whose angle of sharpness is a, 
and JB the block, on wliich rests a weight 
(not shown) whose pressure on B is verti- 
cal and = Q. The weights of wedge and 
block are neglected. There is supposed to 
be no friction, otherwise \\z results would 
be quite different (see § 9 of Graph. Stat. 
of Mechanism, in this book, and p. lYl, M. of E.). Given Q 
and a^ what force P must be applied horizontally at the head of 
the wedge to prevent the block B from sinking (or to raise the 
block with constant velocity if the latter has an upward motion)? 

Supposing the required force P to be in action, and that there 




Fig. 45. 



40 



NOTES AND EXAMPLES IN MECHANICS. 




Fig. 46. 




1= 



Fig. 47. 



is no friction, the mutual pressure, iV, between block and wedge 
is normal to the surface of contact and hence 
makes an angle a with the vertical, while the 
pressures on surfaces cd and eh^ viz., S and R^ 
are horizontal and vertical, respectively. Hence, 
when the block B is considered free (Fig. 46) we 
r?ve equilibrium between the three forces, Q^ iV, and S\ the 
two last are unknown, but are determined thus : 

From ^ (hor. comps.) = 0, /S' = iV^ sin «f ; 
From ^ (vert, comps.) = 0, iV^ cos a: = Q, 
Similarly, we have the wedge free in Fig. 47 under the 
action of iV, now known, and the unknown R 
and the required P. Hence 

R = iV'cos a, {= Q\ from ^¥=0:, 
and P =z ]}^ ^m a, — Q tan a^ from ^X = 0. 

Evidently, the sharper the wedge the smaller 
the force R necessary for a given Q. 

40. Cantilever Frame (Fig. 48). — This frame consists of eleven 
straight pieces in a vertical plane, 
pin-connected, and supposed without 
weight. Pieces C, I), E, and / are 
horizontal, G and H vertical, the 
others oblique. The vertical rod at 
A is anchored, as shown, but no stress 
is induced in it (nor in any other 
piece) unless a load P is placed at w 
(no other joint to be loaded). Since, 
^^^•48- furthermore, no piece connects more 

than two joints, each piece is a straight tioo-foree piece^ subjected 
to a tensile or compressive stress along its axis, and any one may 
be conceived to have a portion removed, when desired, in the 
isolation of the various "free bodies" to be considered. The 
necessity of the rod and anchorage is evident from the fact that 
the rectangle mlnom. is left unbraced. The load P being given, 
it is required to find the stress in each piece of the frame and in 
the rod at A, 




WEDGE — CAiq^TILEVER FRAME. 



41 



The free body in Fig. 49 enables us to find the stresses E 



and F. Since only three forces are concerned, 
meeting at a point, a simple procedure is to 
resolve the known P into two components, E' 
and F\ along the action-lines of E and F^ as 
shown by dotted lines. E' and F' are equal 
and opposite to the required stresses E and F^ 

a 



respectively ; i.e., E =^ P cot a =^ P 



h 



and 




.*!' 



Fig. 49. 




F=P cosec (x= P -r- Bin a. (The summation of horizontal and 
vertical components put equal to zero would give the same result.) 
^'is tension; F, compression. 

Next take the free body in Fig. 50, involving the known force 
P and the unknown stresses P^ I, and Z, 
assuming for them the characters implied by 
the pointing of the arrows. Taking moments 
about centre of pin at 0, we have 

Ph — Pa = 0; whence i> = -f p|, 
Fig. 50. and is tension. 

From ^ (vert, comps.), P = Z cos ^, or Z 

is therefore compression as assumed ; while, from ^(hor. comps.) 
= 0, Z- Z> — Z sin /? =: ; i.e., r= + P (tan j3 + cot a), and 
is compression. 

To find the stresses in A, P, and G we use the free body in 
Fig. 51. By moments about 0, 

Ad-Ph = 0; 

, • . A :=-}- P -^^ and is tension. 
From 2 (hor. comps.) = 0, 



H 7f y and 

' COS p ' 



D — B cos y ■■ 

whence j5 = + t * •» 

' A cos ;/ ' 



0; 




Fig. 51. 



and is compression, as assumed. 




42 NOTES AND EXAMPLES IN MECHANICS. 

In Fig. 4^ we note that since pieces C and D are in the same 
straight line and G is the onlj other piece 
connecting with the joint -m' (there being also 
no load on ???'), the stress in the piece G must 
be zero and the stress in G must equal that in 
D ; similarly, the stress in piece H is zero and 
E = D (as already found above). Hence G is 
marked = in Fig. 52, showing a free body 
Fig. 52. in which the stresses J and K are the only 

unknown forces, G being = Z>, — Pj^ and G = zero. There- 
fore, by elimination between the equations 

J cos 6 -^ G — I — K cos (5^ = (from sum of hor. comps.), and 
J sin 6 -\-K ^m 6 — A = (from sum of vert, comps.), 
we obtain the stresses J and K, both of whicli are assumed to be 
compressions in this figure. 

If the joints on' and n were not both in the horizontal line 
joining w and -y. Fig. 48, the stresses in G and H would not be 
zero, as they are in this instance. 

40a. Real and Ideal Forces. " Balanced Forces." — The student 
should be careful to distinguish between real and ideal forces. 
If a body A receives pressures P and Q from bodies B and (7 re- 
spectively, the resultant of P and Q is purely ^W^aZ, being merely 
conceived to take the place of P and Q^ if any useful and legiti- 
mate purpose can thereby be served. Again, the X and Y 
components of an actual, or real^ pressure P are ideal.^ serving a 
mathematical purpose only, when we suppose P to be removed 
and its components inserted in its stead. 

Such customary phrases as "balanced forces," "forces in 
equilibrium," etc., are unfortunately worded, as they seem to 
imply that forces act on forces, which is an absurdity. In reality, 
"bodies act on hodies, force being the mere name given to the 
action (if it is a push, pull, etc.) ; so that, instead of stating that 
" the forces are balanced," we should more logically say : " The 
rigid body is in equilibrium, or balanced, under the actions of 
certain other bodies." (See correspondence in the London 
Engineer of June, July, and Aug., 1891.) 



CHAPTEK III. 
Motion of a Mateeial Point. 

41. Velocity and Acceleration. — Any confusion between the 
ideas of velocity and acceleration is fatal to a clear understanding 
of the subject of motion. Just as velocity may be defined as the 
rate at which distance is gained, so acceleration may be defined 
as the rate at which velocity is gained, thus : If at a certain 
instant a material point has a velocity of 10 feet per second, we 
mean that it has such a speed of motion that it would pass over 
a distance of 10 feet if that rate of speed remained constant dur- 
ing the whole of the next second of time. Similarly if, at the 
same instant, the acceleration of the material point is said to be 
20 feet per '^ square second " (or '' 20 feet per second, per second "), 
we mean that if the velocity were to continue to change, during 
the w^hole of the next full second, at the same rate at which it is 
changing at the instant mentioned, the velocity at the end of that 
second woidd be greater by 20 units of velocity than at the in- 
stant mentioned, i.e., the velocity would be 30 feet per second. 
According to this definition, then, if the velocit}^ of the material 
point remains unchanged, the acceleration is zero at every point 
of the motion. 

At each point, then, in the path of a material point moving 
along a right line, we have to do with 

5, its distance from some convenient origin in that line ; 

ds 
V, its velocity, or rate at which that distance is changing, = -, , ; 

p^ its acceleration, or rate at w^hich the velocity is changing, 

dv _ d^'s 
^dt "^ 5?" 

There is no need of defining still another quantity as the rate 
at which the acceleration is changing, for the reason that the 

43 



44 NOTES AND EXAMPLES IN MECHANICS. 

force which at any instant occasions the peculiarity of the motion 
of the material point is determined from the acceleration, viz., 
from the relation yc;rc^ == mass X accel. 

Ill the uniformly accelerated motion of a free fall in vacuo^ 
the following are the values of the above quantities at beginning 
of the motion, and at the end of each full second thereafter, thus : 
5, = Distance, d, ■= Veloc. p. — Acceleration = g. 
At beginning, ... 0.0 ft. 0.0 ft. p. sec. 32.2 ft. p. (sec.)'^ ; 
At end of 1st full sec, 16.1ft. 32.2 " " 32.2 " 
At " 2d " " 64.4 ft. 64.4 '' " 32.2 " " 
At '' 3d " " 144.9 ft. 96.6 " " 32.2 " " 

A recent English writer is so desirous that acceleration shall 
not be confused with velocity that he calls the unit of velocity a 
'' jSpeed " and the unit of acceleration a " Hurry?'' For example, 
using the foot and second as units, he would say that at the end 
of the second full second of a free fall in vacuo the velocity is 
64.4 " sjpeeds^'' while the acceleration at the same instant is 32.2 
''^liuinnesP These words are quite suggestive and should be 
borne in mind. 

To say that at a certain instant the acceleration is zero does 
not imply that the body is not moving, but simply that its veloc- 
ity, however small or great, is not changing ; and again, the 
statement that the velocity is zero at a certain instant does not 
imply that the acceleration is zero also, but only that the velocity, 
as its value changes, is then passing through the value zero, while 
the rate at which it is changing (i.e., the acceleration) is not de- 
termined until some further statement is made. 

42. Momentum. — This word is used quite largely in some 
works on Mechanics, but may be considered a superfluity, liable 
to give rise to confusion of ideas ; though sometimes useful, it 
need rarely be used. By definition, it is a name given to the 
product of the mass of a material point hy its velocity at any 
instant^ i.e., Mv. Of course, the mass of the body is a constant 
quantity, while its velocity may be continually changing ; hence 
the momentum is always proportional to the velocitj^ In 
accordance with this definition, the value of a force which is 
accelerating the velocity of a material point in its path is some- 



MOMl^JNTUM — CORD AND WEIGHT. 45 

times stated to be equal to the rate at which the momentum is 
changing • by which is simply meant the following : 

From eq. (lY), p. 53, M. of E., we have P = Mp, = {mass X 

accel.) ; but j9 = -j- ; hence we may write 



dv Mdv d\_Mv\ _ Fchange of niomen- 
dt ~ dt ~ dt L turn in time dt 



-^ dt"^ 



i.e., P — the rate at which the momentum is changing. There- 
fore the above statement as to the value of the accelerating force 
is nothing more than what we already have in the form of 
P = Mp, 

43. Cord and Weights. — There are few mistakes more common 
than rusliing to the conclusion that the tension in a vertical cord 
to which a weight is attached is equal to that weight. This may 
be true (and is true if the weight is at rest and has no other sup- 
port), but should not be assumed without thought. 

For example, Fig. 53, having two weights attached to the same 
cord, if the point A of the cord were fastened, /^ 

the tension in B would be = 6^ : and that in 







u 



a- 



(7, == G'^ if Cr and G' are the- respective 
weights of the two bodies. But if, the cord 
being continuous and not fastened, with no 
friction at the pulley-axles, an accelerated 
motion begins (assume G' > G\ the tension 
on each side depends on that acceleration^ fig. 53. 

as ivell as on the weights of the hodies. To find this accel- 
eration, which is common to both sides, neglecting the masses and 
weights of the pulleys and cord (by which we mean that the 
tension S in the cord at A may be taken equal to that at P and 
C) let us consider the body G' free. We note that this body has 
a downward accelerated motion, and that the forces acting on it 
are G\ directed vertically downward, and S, the tension in 
the cord, pointing vertically upward, i.e., acting as a resistance ; 

hence, calling the acceleration p, we have G' — S = — jp. As 

for the other weight, it is rising with an upward acceleration = z>, 



46 NOTES AND EXAMPLES IN MECHANICS. 

under an upward force S and a downward resistance G, whence 
S — G = —p. From these two relations we obtain by elimina- 
tion 

G'- G , ^ 2GG' 

The value of j? might have been obtained directly by consid- 
ering that the acceleration of the motion is just the same as if the 

whole mass ^ were moving m the same right line under 

the action of a single accelerating force G' — G. 

44. Example. Lifting a Weight. — A rigid mass weighing 100 
lbs. is to be lifted vertically through a distance of 80 ft. in 4 sec- 
onds of time, and with uniformly increasing velocity, from a 
condition .of rest (i.e., veloc. = 0). What tension must be main- 
tained in a vertical cord attached to it, to bring about this result? 

The average velocity is 20 ft. per second, and since the initial 
velocity is zero and the rate of inci'ease of velocity (i.e., the 
acceleration^ p) is to be constant (uniformly accelerated motion), 
the final velocity will be double the average, viz., 40 ft. per sec. 
(see also eq. (4), p. 54, M. of E.). If, then, 40 velocity-units are 
gained in 4 seconds, the acceleration is 40 -^ 4 = 10 ft. per second 
per second (10 " hurries''), and an upward accelerating force of 

Mp = ^^-^ X 10 = 31 lbs. must be provided. But the only 

forces actually present, acting on this mass, are the action of the 
earth, viz., 100 lbs. pointing vertically downward, and the ten- 
sion, xS, in the cord, directed vertically upward ; and their (ideal) 
resultant, which is /i^ — 100 lbs. (since they have a common action- 
line), is required to be 31 lbs. and to act upward; whence we 
have the required cord-tension, = xS, = 131 lbs. 

To lift the 100-lb. weight, then, under the conditions imposed, 
requires a cord-tension of 131 lbs. [If the cord be allowed to 
become slack at the end of the 80-ft. distance, the body does not 
immediately come to rest, as it then has an upward velocity of 
40 ft. per second. Its further progress is an '' upward throw" 



HAKMONIC MOTION". 47 

(p. 52, M. of E.) with 40 ft. per second as an initial velocity, the 
only force acting at this stage being the downward attraction of 
100 lbs., which will gradually reduce the velocity to zero.] 

If a smaller height of hft had been assigned with the given 
time, or the same height with a longer time, the difference be- 
tween the requisite tension S and the gravity force (or weight) of 
100 lbs. would have been smaller ; and vice versa. 

It is convenient to note that in using the foot-pound-second 
system of units for force, space, and time, the mass of a body is 
obtained by multiplying its weight in pounds by 0.0310 (which 
is the reciprocal of 32.2) ; thus, in the preceding example the 
mass of the 100-lb. weight is = 100 X .0310 = 3.10 mass-units. 

45. Harmonic Motion. — From eq. (3), p. 59, M. of E., remem- 
bering that c and a are constants, we note that s, the " displace- 
ment," or distance of the body from the origin (middle of the 
oscillation), is proportional to the sine of an arc or angle, and that 
this arc is proportional to the time, t, elapsed 
since leaving the origin ; whence arise the 
following graphical relations : In Fig. 54, let 
be the origin or middle of the oscillation, 
and the horizontal line CD the path of the 
body ; OD, = r, being the extreme displace- 
ment, i.e., the " semi-amplitude," of the mo- 
tion. At any instant of time the body is at 
some point m between C and D, i.e., Om = the variable s (dis- 
placement). With as centre, and OD, = r, as radius, describe 
a circle and erect a vertical ordinate at m at whose intersection n 
with the curve draw a radial line 0?i, making some angle 6 with 
the vertical, and the abscissa nk. Now OiJi, or s, ^ r sin 6 ; and 
the length of curve £'n is proportional to the angle 0. Hence 
the linear arc En is ^proportional to the time occupied hy the 
'body in describing the distance s = Om ; that is, if n be re- 
garded as a moving point, confined to the circumference of the 
circle, and always in the vertical through m, its motion being 
thus controlled by the harmonic motion of m, the velocity of n in 
its circular path is constant (equal linear arcs in equal times). 
The constant velocity of n must be equal to the initial velocity 




48 NOTES AND EXAMPLES IN MECHANICS. 

of m, i.e., <?, as the latter leaves the origin, 0\ for at the two 
points are both moving horizontally. 

Conversely, therefore, if a point n move in the circumference 
of a circle with a constant velocity c (continuously in one direc- 
tion), the foot of its ordinate, i.e., the point m^ moves with har- 
monic motion along the horizontal diameter ; this is the proof 
called for^in the line below Fig. 64 of M. of E. The d of Fig. 
54 is t Va of the formulae on p. 59, M, of E. 

Since the acceleration (as also accelerating force) in harmonic 
motion is proportional to the displacement, the length of the line 
Orn, or Jen, represents the acceleration at any instant, while the 
length, mn, of the ordinate is proportional to the velocity of the 
body or material point, m (since by eq. (4) of p. 59, M. of E., 
that velocity varies as the cosine of 6). 

Of course, in obtaining values, from the drawing, of these 
variables v, p, and z5, for different positions of the body m, due 
regard must be paid to the scales on which the lengths marked 
in the figure represent these variables, none of which is a linear 
quantity. 

46. Numerical Example of Harmonic Motion (using the foot, 
pound, and second). — With the apparatus and notation of p. 58, 
M. of E., suppose that by previous experiment with the elastic 
cords we find that a tension T^ of 4 lbs. is required in either of 
them to maintain an elongation of 3 in., i.e., i of a foot ; then 
for any elongation s (or '' displacement" of the body during its 
motion) the tension (and retarding force) is T = T^s -i- s^ = 16s. 
Let the small block weigh 128.8 lbs., then its mass is 128.8 X .0310 
= 4: = M ; and hence the constant quantity called a is 
a = {T,~Ms,) = 16-^M; i.e., a =4.. 

Let an initial velocity of c = 4 ft. per sec, from left to right, 
be given to the block at ; required the extreme distance attained 
by the block from (i.e., the semi-amplitude," = r), and the time 
occupied in describing the semi-amplitude (i.e., required the time 
of a quarter-period). 

From p. 59, M. oi E., r = c -~ V^; .: r = 4: ~- VI = 2 it; 
while for a quarter-period, or half-oscillation, we have the time 
^Tt -r- Va = iTt = 0.7854 sec. 



BALLISTIC PENDULUM. 49 

Hence in the diagram of Fig. 54 we make OD = r = 2 it. 
Then, since the linear arc £^D represents 0.7854 seconds, the time 
represented by mn for any position m of the block will be on a 
scale of ^ sec. to each foot of Mi (since 0.7854 -^ {^rr x 2 ft.) = \). 
Since OE, or 2 ft., represents the velocity of m on leaving O^ 
i.e., 4 ft. per second, mn represents the velocity at m on a scale 
of 2 ft. per second to each foot of mii. Similarly, since the 
acceleration OD of m at D (from j9 = — <25) is — (4 X 2) = — 8 
ft. per (sec.)", that at m will be ku on a scale of 4 to each foot of Icn. 

It must be noted that after m- reaches D and begins to return 
toward the left, the point n in the curve will have passed helow 
the horizontal through 7> ; that is, the time is now greater than 
ED^ and the velocity has changed sign. When n passes under- 
neath the acceleration and the displacement change sign ; and 
so on indefinitely. (Tension in cord at i> = ?) 

47. The Ballistic Pendulum. — This old-fashioned apparatus for 
determining the velocity of a cannon-ball consisted of a heavy 
body B, of mass M^ , suspended, by a rod 
or rods, from a horizontal hinge on a fixed 
support. A cavity in its side is partly filled 
with clay, so as to make the impact inelastic. 
This body being initially at rest (see Fig. 55), 




the ball is shot horizontally into the cavity, in ^-i |^^ ^^' 
which it adheres. As a result of the impact cif^Oi ~ 
the centre of gravity of the combined masses 
receives a velocity (7, with which it begins its 
ascent along the circular arc JBm, finally reaching a vertical 
height ^ above its initial position before coming to rest (for an 
instant). H being measured, or computed from the observed 
angle /?, we have C= + V^yH (see foot of p. 80, M. of E.). 
The known mass of the ball being M^ and its unknown velocity 
just before impact -|- c^ , eq. (4) of p. 65, M. of E., enables us to 
write 

(In this theory the suspended mass has been treated as a 
material point, and the result is only approximate. The longer 



50 NOTES AIS^D EXAMPLES IN MECHANICS. 

and lighter the suspension-rod and the smaller the dimensions of 
the suspended body the more accurate the results. The strictly 
correct theory is rather complicated.) 

48. Apparatus for the Determination of the ^' Coefficient of 
Restitution." — It will be noted, by an examination of eqs. (2), (3), 
(5), and (6), on pp. 64 and 65, M. of E,, that the coefficient of 
restitution, e^ may be defined as the ratio of the loss of momentum 
of the first body in the second period of the impact (period of 
expansion or restitution) to the momentum lost by it in the first 
period (compression) ; and similarly for the second body, except 
that for it there is a gain of momentum in both periods, instead 
of a loss ; hence 

Fig. 56 shows the apparatus. The two balls, of same sub- 
stance, are suspended by cords so 
^ that they can vibrate in the same 

>, V vertical plane. When hanging at 

\ \ rest they barely touch, without press- 

\ \ jy ure, with centres at the same level. 
\b, '' Being allowed to swing simultane- 
^^--^|h2 1 ■ ously from rest at <x and h respec- 
^7.lt)Z.JrQ^^Z--j^'y- "^ tively, their impact takes place at O 
Fig. 56. (or very nearly so), their velocities 

just before impact being <?i = + V^gh^ and c^— — V2gJ\ , re- 
spectively (see foot of p. 80, M. of E.), where A, and h^ are the 
vertical heights fallen through (each ball having underneath it a 
graduated arc, so that each of these heights can be computed 
from the observed angle and known length of cord). Supposing 
each ball to rebound on its own side of O and the heiglits reached, 
H^ and^2, to be noted or computed, their velocities at O im- 
mediately after impact must have been F, = — V^gH^ and 
y^=z -\- V 2^ /^respectively, A and B being the points reached 
at the ends of the rebounds. Knowing, then, h, and A^? ^i and 
M^^ II, and 11^, we compute e from either eq. (7) or (8) of p. 65, 
M. of E. 




SPRING BETWEEI^ TWO BALLS. 51 

Of course e is always less than unity, and due regard must be 
paid to the signs of the velocities in making the computations. 

49. Action of a Spring between Two Balls (Fig. 57). — In the 
second period of a direct central impact 
(period of expansion, or restitution) the dis- 
tance between the centres of the two masses 
is increasing, the front body being subjected 
to a forward pressure precisely equal to that 
which at the same instant is retarding the 
hinder body. Hence the gain in momentum ^— "__ ^^^^^ .__ t_L^ 
of the front body during this period is fig. 57. 

equal to the loss in momentum of the other body ; i.e. (see eqs. 
(5) and (6), p. 65 of M. of E.), Ml Y, - C) = MIC - V,). 

This is precisely what takes place when a spring in a state of 
compression (the ends tied together by a cord) is placed endwise 
between the two suspended balls of the apparatus of Fig. 57, and 
the cord is then severed without violence (burning is best). The 
spring in regaining its natural length exerts equal horizontal 
pressures in opposite directions at any instant against the two 
masses. This pressure is variable, but at any instant has the same 
value for one mass as for the other. That is, this phenonienon 
may be looked upon as a case of impact where the first period is 
lacking, the period of restitution occupying the whole time of 
impact. In this case M^ and 21^ are at rest just before impact, 
that is, (7=0; while their velocities after the expansion of the 
spring are — V^gH^ and -|- V^gll^ respectively (if H^ and H^ are 
the respective vertical heights reached by the balls as the result 
of the action of the spring). Hence, from the above equation, 
J/,(+ V'2gH, — 0) = JfXO — [— Vigll'^) ; i.e., the velocities 
immediately after the action of the spring are inversely propor- 
tional to the masses. 

The same method in another form consists in saying that from 
the principle of the conservation of momentum the total momen- 
tum before impact, viz., 0, here, is equal to that after impact, 
which is M, V, + M, F,, or MX - V2^) + Ml + 4/2^). 
Equating the latter expression to the zero momentum first men- 



52 NOTES Al^D EXAMPLES IN MECHANICS. 



tioned, we have 31, V2gII, — M^ V^gH^ , as before. This result 
is verified bj the apparatus. 

50. The Cannon as Pendulum. — As the converse of the Ballis- 
tic Pendulum, the cannon itself may be suspended in a horizontal 
position from a pivot C\ Fig. 58. With the ballistic pendulum 
^^j^ the impact was inelastic, i.e., it consisted of a 

period of compression only, that of restitution 
being lacking. Here, however, the expansion 
of the hot gases generated when the powder 

y^'l B is ignited acts like the spring in the preceding 

^^^_^^,,^^ivi2 case, causing a forward pressure at any instant 
"^"^^Cii-^*"^ against the ball and an equal and simultaneous 
Fig. 58. backward, pressure against the cannon ; i.e., 

the impact is one having no period of compression but only one 
of restitution. The velocity V^ of the ball leaving the muzzle 
may therefore be inferred from a knowledge of the two masses 
concerned and the height of recoil H, {A being the point where 
the centre of gravity of the gun comes to rest for an instant). 
Putting the total momentum before impact, which is zero (cannon 
at rest, in lowest position), equal to that immediately after, the 
cannon having then just begun its motion from D toward A with 
a velocity at i> of T^ = — V^gR, , we have 

M - 

= Jf,F, + J/X-l/2^^:); or Y, = ^-V2gH,. 

As before, we here treat the masses as material points. 

61. Simple Circular Pendulum of Small Amplitude (Fig. 59). — 
^ is a material point suspended from a fixed 
point C by an imponderable and inextensihle 
cord. Being allowed to sink from initial rest at 
the point A (cord taut), it follows the circular 
arc ABO with increasing velocity. At any 
point B between A and its velocity (which 
of course is tangent to the curve) is 

V = V2g X vertical height Z>i^,' 
(see foot of p. 80, M. of E.), and the forces acting on it are its 
weight G, vertically downward, and the tension S, in the cord. 
This tension increases as the body approaches O, since 2 (norm. 




SIMPLE ^IRCULAil PENDULUM. 53 

comps.) must = Mv'' -^ rad. of curv., [eq. (5), p. 76, M. of E.], i.e., 

cos + ^7 J- • (1) 

r 9> 

At 6^ the tension is greatest and = 6r 



G v' 
S — G cos (p ^= — • -r : whence S = G 
g I ' 



f^]. If the 

starting point A is taken high enough, the cord maj be broken 
before is reached. To find tlie tangential acceleration pt, or 
rate at which the velocity is increasing, we note that from eq. (5), 
p. 76, M. of E., 2 (tang, comps.) must = Mpt, whence 

Mpt = G sin (p + zS XO; or ^^ = ^ sin 0. . (1) 

This can be written p^ = ^ ■ I sin 4> = j- • ^T". ...» (2) 

But if the angle ACQ is very small (not over 5°, say), BF 
may be put = linear arc OB ; and as is the middle of the 
oscillation, and BO is the distance or displacement of the body 
from at any instant, (2) may be stated in the form : the accelera- 
tion is projportional to the disjplacement \ so that the motion is 
(very nearly) harmonic (see p. 58, M. of E.). To find the time of 
passage from ^ to on this basis (half-oscillation), note that on 

7t 1 

p. 59, etc., M. of E., the time of a half-oscillation is ^ • — -. 

"^ Va 

where a is the quantity which multiplied by the displacement s 
gives the acceleration. Hence from eq. (2) above, the " (x" of the 
present harmonic motion is g ^l. Hence the time of a whole 
oscillation from A to the corresponding extreme point on the 
other side of is 



'-'/r 



The duration of the oscillation is independent of the amplitude 
(with above limitations). (For a large amplitude see foot of p. 
81, M. of E.) 

With an extensible cord, elastic or inelastic, the results would 

be quite different. In such a case the relation v = y2g . DF for 
the velocity at any point B would not be true, since the ten- 
sion 8 is not perpendicular to the velocity when the cord is 
elongating. 



CHAPTER lY. 

Numerical Examples in Statics of a Rigid Body and 
Dynamics of a Material Point. 

52. Example 1. Anti-resultant of Two Forces (Fig. 60). — Two 

Nr-3Q°j forces in a horizontal plane, J^ and P\ 

P 3 tons' 1 ' L J 7 

pLd tons ' ^' —^ ^^^ respectively 3 tons (5A6>ri5 tons, of 2000 

1 /%^,/^/ lbs. each), and 6000 lbs. P is directed IsTorth 

w oj^lL4,. ^^° ^^^*' ^^^ -^'' ^^^*^ S^° East. The 

^V^r^ ^•'■- ane^le between them is therefore 65°. Re- 

j/^^e^^ ' quired the amount and position of R\ their 

f^' *""!_ anti-resultant (i.e., the force that will bal- 

Fig. 60. ancc them). R' must be equal and opposite 

to the (ideal) resultant, B,^ of the two forces. Adopting the short 

ton as a unit for forces, we note that P' and P are equal, each 

being 3 tons. Hence the parallelogram of forces formed on them 

as sides is a rhombics, and P bisects the angle between ; whence 

we may write 

P=^P cos 32i° = 2 X 3 X 0.8434 = 5.06 tons. 

Therefore an anti-resultant of 5.06 tons, directed South 62^° West 
must be provided, if the two given forces are to be balanced by 
a single force. 

53. Example 2. Resultant Couple of a Number of Couples. — 
Six couples acting on a rigid body and in the same plane (a ver- 
tical plane) have the following moments, respectively, as seen 
from the west side of the plane : 

+ 30 ft.-lbs. ; + 196 f t.-oz. ; + 0.48 inch-tons ; 
— 0.08 ft.-tons ; - 160 ft.-lbs.; and — 1300 inch-lbs. 

Required the moment of their resultant couple. 

54 



NUMERICAL EXAMPLES — STATICS, ETC. 55 

Changing the form of these various moments to what they 
would have been if all the various forces had been expressed in 
lbs. and the arms in feet (i.e., expressing them all in ft.-lbs.), and 
adding those of the same sign, we have 

+ (30,+ 12.25 + 80) ft.-lbs. = + 122.25 ft.-lbs. ; 
and - (160 + 160 + 108.33) ft.-lbs. = - 428.33 " 

-306^ " 

Since the algebraic sum of the moments is — 306.08 ft.-lbs. 
(the couples being in the sa?ne plane (see § 34, M. of E.), this is 
the moment of their (ideal) resultant couple. To hold the given 
couples in equilibrium, therefore, a sitigle couple (their anti- 
resultant) having a moment of + 306.08 ft.-lbs. must be applied ^ 
to the body to preserve equilibrium. That is, if a seventh couple 
in which (for example) each force is 153.04 lbs., with an arm of 
2 ft., and appearing counter-clockwise seen from the west, be 
applied to the body, the seven couples will balance. 

If the given couples were not all in the same plane, or in 
parallel planes, we should have to make use of the results of § 33, 
M. of E. 

54. Example 3. Centre of Gravity of Trapezoid with Semicircle 
Cut Out (homogeneous thin plate of uniform thickness — Fig. 61). — 
The diameter of the semicircle lies on | 
the lower base OJK of the trapezoid, J3^ y /'^" 
is the centre of gravity of the semi- L 
circle (so that IID =z 4:r ^ Stv), E that 
of the complete trapezoid, and E' that 
of the actual plate. See figure for 
given numerical dimensions and for 
notation of co-ordinates of centres of 
gravity of actual plate and complete fig. 61. 

trapezoid, denoting those of the semicircle by x^ and y^ . Re- 
quired x^ and y^ , using the inch as linear unit. The area of the 
trapezoid is i^ = ^ X 6 X (10 + 16) = 78 sq. in. ; that of the 
semicircle is F, = k^ti^Y = 6.28 sq. in. ; so that the area of the 
actual plate is their difference, F^ , =: 71.71 sq. in. Also, for the 
point II we have y^ = 4 X 2 -f- 3;r = 0.85 in. ; and OD = x^ 

* In the same, or iu a parallel, plane. 




56 



NOTES AND EXAMPLES IN MECHANICS. 



= 8 in. ; while (see p. 23, M. of E.) y = ^ 

o 



6 16 + 20 



= 2.77 in.; 



16 + 10 

and x= OD — ^X 2.77 = 7.54 in. 

{C bisects the base JI/iT; conceive perpendiculars let fall 
from C and ^ upon OK, and use the similar triangles so formed 
in obtaining the value of x.) 

From eq. (3), p. 19, M. of E., since the trapezoid is made up 
of the actual plate and the semicircle, we have 
x = {F,x, + F^x^)-^{F,-\-F,)', 



Fx-F,x, 



and 



2/i = 



and 



Equal lengths. 



Bj substitution, therefore, 

\ = [78 X 7.54 - 6.28 X 8] -^ 71.71 = 7.5 in. ^ 
y, = [78 X 2.77 - 6.28 X 0.85] -^ 71.71 = 2.94 " 
55. Example 4. Stability of Two Cylinders. — Fig. 62 gives an 
Two Cylinders, end vicw of two smooth and ho- 
mogeneous circular cylinders of 
equal length (= Z' in feet) but of 
radii 1 in. and 3 in., respectively. 
A weighs 800 lbs. per cub. ft., 
B only 100 lbs. per cub. ft. A 
and £ being placed, as shown, on 
two smooth planes at 45° with the 
horizontal, it is required to find 
whether this is a stable position. 
Call their total weights G, and G, 
Since OD is one half of TJO, the angle COD is 
First, the position being supposed stable, to find the press- 
ure at point 2, we take JL as a free body. The forces acting on it 
are three, shown at in Fig. 63. P^ is the pressure (or reaction) 
of the inclined plane against A at point 1, P^ is the pressure 
from the other cylinder, and G, the weight of this cylinder, A. 
These are all directed through (smooth surfaces), the angles 
being as shown. For equilibrium F^ must be equal ai.d opposite 
to OJT, the (ideal) resultant of G, and P, . In triangle OP^K^ 
0K\ G, :: sin 45° : sin 60° ; .-. 0^= G,l V2 ~ 4/3 ] = P, . 




Fig. 62. 

or if JL will crowd B out of place. 

for the present. 

30°. 



NUMERICAL EXAMPLES— STATICS. 



57 



/' 






I 



Fig. 63. 



'-:h 



In Fig. 63 we have acting through G the four forces acting 
on the large cyhnder B (on supposition _ ^Pi 

of equihbrium). G^ is its weight, P^ is 
the pressure of the other cylinder, P^ 
the pressure at point 3 of the inclined 
plane on the right, P^ of that on the 
left. 

If now the resultant of P^ and G^ were found to pass above 
the point 3, instability would be proved; since to occasion 
pressure at point 4 that resultant should evidently pass below 3. 
Or, which amounts to the same thing, assuming equilibrium at 
and with P^ as drawn, if we compute the value of P^ by put- 
ting ^ (compons. -| to P^ = 0, and a negative result is obtained, 
instability is proved ; and vice versa. Hence we write 



P, = G, cos 45° - P, cos 60^ 



whence P. = —^ — -^. 

V2 t6 



(1) 



Now G, = l7r{-^\y X SOOl'] lbs. ; and G, = [7r{j\y X lOOr] lbs., 
on substituting which in (1) we obtain 

P, = [^|^;rZX + 309.8)]lbs.; 
which is positive, and thus verifies the supposition of equilib- 
rium. 

56. Example 6. Toggle-joint (Fig. 64). — The two straight 
links, of equal length, are pivoted to the two 
blocks, as indicated. A horizontal pull of 80 lbs., 
making equal angles with the two links, is exertejj 
on the horizontal pin of the joint P. What press- 
ures are thereby induced (for given position of 
parts) on the surfaces P and P (horizontal and 
vertical)? (Those on P' and P' will be the 
same, respectively.) Each link is evidently a 
straight two-force piece and hence under a com- 
pressive stress along its axis ; call this stress P\ 
The free body in Fig. 65 enables us to find P^ 
from ^ (hor. comps.) = ; i.e., 2P^ cos «r = P ; 




orP' = 



2 cos a 



58 



NOTES Al^D EXAMPLES IIST MECHANICS. 





Fig. 65. 



Pig. 66 shows the upper block free, from which bj ^1^= 0, 

Pe = P^ sin a, and from hor. 
comps. Pjr = P' cos a. 

.'. Pe=^P tan «'=8/.%4^480 lbs. ; 

and Pj.= iP:=40 1bs. 

If, in Fig. 65, P were resolved 

into components along the axes of 

the two links, each such component 
■■' B would be the equal and opposite of 
Fig. 66. the Corresponding P' , Evidently, 
if oi approached a right angle, P\ and also P^, w^ould increase 
without limit. 

67. Example 6. Simple Crane. — The simj^le crane in Fig. 67 
carries a load of 4 tons at C, 
12 ft. from the axis of the 
vertical shaft, while its own 
weight is 1 ton, the centre 
of gravity being 3 ft. from 
the shaft. The socket at B 
is shallow so that lateral 
support is provided at A, 
Required the pressure at A j^^/^j^y 
and the horizontal and ver- ^i®- 67. fig. 68. 

tical pressures from side and bottom of socket B. The crane 
being considered free in Fig. 68, and the reacting pressures being 
put in, as shown, we have, from '2 (moms, about P) = (foot 
and ton), 

J.^X8 — 4x 12— 1X3=0; whence A^ = 6.37 tons. 
From ^ (vert, compons.), P^ — 5 = ; 
-2 (hor. compons.), B^ — A^^:^^\ 




or 



By = 5 tons. 



or 



6.37 tons. 



58. Example 7 Door and Long Hinge-rod. — The door weighs 
200 lbs., and is supported in a vertical plane in the manner 
indicated in Fig. 69. The continuous vertical rod ^^C' is con- 
sidered without weight, and from the nature of the mode 
of support of the door receives horizontal pressures at .each 



NUMERICAL EXAMPLES — STATICS. 59 

of the points J., ^, 6^', and C. Eequired the values (J., B, G\ 
and C) of these forces ; also the vertical pressure C" between 
the projecting shoulders (7 and C, 




A 

\\ 

I 

!o 



=l'6 



lbs. 
G 



tc" 

Fig. 70. 

Fig. 70 shows the door free, and, bj moments about 0\ using 
the pound and foot, 

6^a = ^^ ; or ^ = (200 X 1.5) -f- 5 = 60 lbs. 

-2" (vert, comps.) gives 
C" =G= 200 lbs. ; while from 

2 (hor. compons.), — A-\- C = 0, whence 
C = A^m lbs. 

Tbe rod as a free body is shown in Fig. 71 and enables 
us to find the pressures B and (7, now that A and C are 
known. Bv moments about <?, we have 

^X4:-60x5=:0; whence ^ = 75 lbs. '"'^^ 

In putting 2 (hor. compons.) = 0, since C ^= A 
from Fig. 70, the summation reduces to C—B = 0, 
i.e., 6^= 75 lbs. 

The hinge-rod, therefore, is seen to be under 
the action of two couples of equal and opposite 
moments ; one consisting of A and C\ the other 
of B and C, while the sill C^ Fig. 69, receives a vertical pressure 
equal to the weight of the door. 

59. Example 8. Shear-legs (Fig. 72).— The weight G of 



4'=;^ 



6=5' 



Fig. 71. 



60 



NOTES AND EXAMPLES IN MECHANICS. 



G = 2000 Zfcs. 




^-lol^^'^v 



Fig. 73. 



2000 lbs. is supported bj the two straight links in a vertical plane 
as shown, with given dimensions and weights. Required 
the pressures produced on the hinge-pins at A and £. If the 
links had no weight they would be straight two-force pieces, CB 

being subject to a compression 
and AC to tension; and the 
hinge-pin pressures at A and B 
would be equal to these forces, 
the action-li n es of the latter bei ng 
the axes of the pieces, respective- 
ly. In that case a simple solution 
would consist in resolving the 
force of 2000 lbs. by a parallelo- 
gram into two components, oue along CB, the other along AC 
prolonged, and these components would be found to be 3367 and 
1813 lbs., respectively; the former being the compression in CB 
and the latter the tension in CA. But the weights of the links 
are considerable and are to be considered ; hence the links are 
not two-force pieces and must not be conceived to be cut in form- 
ing any free body. The hinge-pressures at A and B are replaced 
by their horizontal and vertical components, as shown, A^ and 
Ay , B^ and By ; these four are the unknown quantities re- 
quired. 

The figure shows all the forces acting on a free body consisting 
of the two links and the 2000-lb. weight. By taking moments 
about A we exclude three of the unknown quantities and 
obtain 

+ 2000 X 30.+ 600 X 26 + 900 X 16 - ^^ X 20 =: ; 

or By = 4500 lbs. ; 

while by moments about B, 

2000 X 10+ 600 X 6 - 900 X 4 — ^j, X 20 = ; 



or A, 



1000 lbs. 



Now conceive the link CA alone to constitute a free body, 
the forces acting being A^, Ay, G^, and the pressure of the 



NUMERICAL EXAMPLES— STATICS. 



61 



hinge-pin at C against this link. The action-line of this last 
force is not known, but the moment-sum about C excludes the 
force and gives 
+^^ X 20 — 1000 X 30 -^ 900 X 14 = ; whence A^ = 2130 lbs. 

Since in the first free body £^ must = A^ (from 2 hor. 
comps. = 0), we have also B^ = 2130 lbs. and can now compute 
the actual oblique hinge-pressures A and ^, at A and £ re- 
spectively. From 

A = VA: + a;, and B = V^^ + By% 
we have finally 

^ = 2354 lbs., and ^ = 4978 lbs. 

From tan " \Ay -i- A^) we find that A makes a smaller angle with 
the horizontal than the link CA ; and similarly, that of £ is 
greater than that of link CB, 

60. Example 9. Roof Truss with Loads and Wind Pressures 
(Fig. 73). — Here the half- weight of each piece is supposed to be 




2'tons 



Fig. 73. 

carried directly on the pin of the corresponding joint, so that 
each link or member will be considered as a ^irdii^ht two-force piece 
and hence in simple compression or tension along its axis. In 
obtaining free bodies, therefore, any piece or pieces may be con- 
ceived to be cut and the stress inserted. The load given at each 
joint includes the half-weisrhts of all the pieces meeting there. 
For all distances and angles needed see figure. The wind is 
supposed to blow from the left, its pressure (4 tons) on the left 



62 NOTES AND EXAMPLES IN MECHANICS. 

slope of the roof being normal to the same, half borne at each 
joint, h and c. Resistance to horizontal displacement is supposed 
to be provided at the right support, alone ; the other extremity 
of the truss being on rollers, so that the reaction there is vertical ; 
hence at the right we have two reactions to deal with, horizon- 
tal and vertical ; i.e., Hy^ and F^. 

Bequired the three supporting forces, H^^ , Y^ , and Y^ ; and 
also the stresses A, (7, i>, and £ (and their character), in the 
pieces, A^ C^ D, and E. 

Fig. 73 shows the whole truss as a free body. By moments 
about joint Z>, adopting the foot and ton as units, we have 

+ F, X 38 - 2 X 38 - 3 X 12 - 3 X 26 -4x19 - 2 X 17.3 = 0; 
whence F^ = 7.91 tons. From ^ (vert, comps.) = 0, 

"^0 + F^ - 2 - 3 - 3 - 4 - 2 - 2 sin «' - 2 sin A == ; 

and hence Y^ = 8.88 tons ; while 2 (hor. comps.) = gives 

+ 2 cos «: + 2 cos «^ — ^„ = 0; or ^„ = 2.86 tons. 

Next, considering free the portion of the truss on the left of 

|3 tons. a plane cutting pieces A^ D^ and C^ we 

jfonsX^c ^^c^ have Fig. 74, in which, for the present, 

p\^ we assume A to be tension and C and D 

, compression. '2 (moms.) about point c 

{intersection of C and D) gives 
^ ^X10.2+2X12+2X17.3-8.88X12=0; 
whence J. = + 4.7 tons, and this being 
positive, the assumption of tension is con- 
firmed. 2 (moms.) about a, similarly, gives 
6'X9 + 3X7-2X1.7 + 2X 19 + 2 X 15.6 - 8.88 X 19 = 0; 
or, C := -\~ 9.1 tons, and is therefore compression. Although the 
same free body would serve, let us determine stress D from 
another free body, that in Fig. 75, showing the remainder of the 
truss. Assume D compression. 

From ^ (vert, comps.) =: we have 

+ 7.91 -3—2-4- J.sin/?-Z>cosr = 0; 
or, D = — 2.45 tons. The negative sign showing the assumption 
of compression to be incorrect, D is 2.45 tons tension. 




NUMEKICAL EXAMPLES. 



63 



Again, from the free body in Fig. 76, taking moments 





Vn= 7.91 tons. 

Fig. 75. 

about n, having assumed ^'to be tension, we have 

+ Z X IT + 3 X 12 - 9.1 X 12.5 = ; 
i.e., ^= -\- 4:.d6 tons, and is tension. 

From the same free bodj the stress in G is easily found. 

61. Remark. — The foregoing examples of this chapter have 
all involved the equilibrium of rigid bodies, each under a system 
of forces in a plane. Those remaining to be given, however, 
deal with moving material points, or bodies small enough to be 
so considered (dynamics of a material point) ; and the concurrent 
forces in each case acting on the body when considered free, do 
not form a balanced system (unless the motion is rectilinear and 
of constant velocity), so that ^X and 2Y are not = neces- 
sarily. For example, if the path is a straight line which is taken 
as the axis X. then ^^= mass X accel. ; while ^ {comps. -\ to 
the path) = 0, as if the forces were balanced. But if the path is 
a €u?'ve, then at each point ^ (co7nj)s, along the tangent) = mass 
X tan. ace. and '2 [comps. along the direction of the normal) 
= Tnass X square of veloc. -h radius of curvature. In numeri- 
cal substitution the student is very apt to forget that if ^, the 
acceleration of gravity, be denoted by the number 32.2, times 
must be expressed in seconds and distances mfeet. (The expres- 
sion for the mass of a body always involves the quantity g) 

62. Example 10. Train Resistance. — If the frictional resist- 
ance of a certain 200-ton railroad train be assumed to be equivalent 
to a backward force of 12 lbs. per ton applied directly to the car- 
frames at any ordinary speed, in what distance on a level track 
will the train be stopped if moving initially at a velocity of 40 



64 NOTES AND EXAMPLES IN MECHANICS. 

miles per hour? (There are no brakes on, nor any locomotive; 
the resistance being due to the rubbing of the journals in their 
boxes and the unevenuess and compressibility of the track and 
wheel-treads (rolling resistance) ). 

Ditto : if the train is on an up-grade of 26.4 ft. to the mile ? 

Taking the axis -\- X m the direction of motion, we note that 
the accelerating force is — 2400 lbs. ; i.e., that the sum of the 
comps. along the path is — 2400 lbs. The mass, in the ft.-lb.-sec. 
system of units, is == 6^ -^ ^ = 400,000 -^ 32.2. ]^ow :SX = Mp 
= 7nass X coco., and therefore the ace, =^p =^J^-^- J!/ == — 0.193. 
The accelerating force being constant, the acceleration is constant 
and hence the motion is uniformly accelerated (retarded here), 
and the eq. (3) of p. 54, M. of E., is applicable, viz, distance — s 
= {v^ — c^) -^ 2j?. The initial velocity = <? = 40 miles per hour, 
= 58.6 ft. per sec, while v is to be zero. Hence 

On the up-grade, the path, or axis X^ is inclined upward at an 
angle a with the horizontal (whose tangent is 26.4 -^ 5280 = g^ 
and is practically = sin a) and, besides the — 2400 lbs., the X 
component of 6r, viz., — 6^ sin a = -^ of 400,000 lbs. = 
— 2000 lbs., acts to retard the motion. 

.•. :EX = — 4400 lbs. and p = — 0.354 ft. per sec. per sec. 
.-. 5 = [0^ - (58.6)^] -^ [2 X (- 0.354)] = 4858 ft. 
63. Example 11. Inclined Plane, Two Weights, and Cord 
(Fig. 77). — The cord connecting the two weights is very light and 
Meightof f^iQibs.-. yf^o inextensible, and friction and mass of the 
Weight of E 12 lbs. y/y L puUcy arc neglected. (By neglecting the 
Tnass or inertia of the pulley we mean that, 
notwithstanding the fact that its rotary mo- 
tion is accelerated by the cord, the tension in 
the cord is the same at any instant where it 
Fig. 77. leavcs, as where it winds upon, the pulley- 

rim.) The two blocks being at rest in the position shown (cord 
taut, with a temporary support under B\ the support is sud- 
denly removed ; required the distance s^ through which B then 





NUMERICAL EXAMPLES — DYNAMICS OF MATERIAL POINT. 65 

sinks in the first two seconds of time (= t^^ and also the tension 
in the cord during the motion, if the body A encounters a fric- 
tional resistance, on the inclined plane, always equal to -f-^ of the 
normal pressure on the plane. 

Consider A free at any instant of the motion (Fig. 78). Call 
its acceleration ^f. The forces acting on 
it are its weight G, S the tension in the 
cord, the friction Jp^, and the normal 
pressure iV^, from the inclined plane. 
(That is, the resultant of JV^ and j?^is the 
resultant action of the plane on body A, 
which resultant action is evidently not normal to the plane, which 
it would not be * unless the bodies were smooth.) Although the 
path of A is straight, consider it as a particular case of a curve ; 
then (see § 61 above) 

/^ 
2 {tang, comps,) = S — F — G co^ a = —p^ ;....(!) 

if 

2 {normal comps.) = ; or N^— G sin a = = 0. (2) 

From (2) we find the value of JV^ ; and hence 

F=^\W,=.^\GB\na (3) 

At this same instant (which is any instant of the motion), con- 
sider B free in Fig. 79. There are only two forces acting on it : 
its weight G^ , and the upward tension S^ in this part of the cord. 
£ is sinking with some acceleration^. 

From 2 {downward comps.) = mass X ct^cc., we have 

G,-S,^fp (4) 

But (from above remark on mass of pulley, etc.) we know that 
/(S'l — xS'; and since the cord is taut and does not stretch, p must 
= Pt . (Let the student devise a strict proof of this.) 

Hence by elimination between the four equations we obtain 

G,-F-G cos a 



* That is, not necessarily. 



66 NOTES AND EXAMPLES IN" MECHANICS. 

which is constant and hence the motion is uniformly accelei ated 
and the equations of § 56, M. of E., hold good, among which is 
s z= ij)f when the initial velocity is zero. 

Passing to numbers, in the ft.-lb.-sec. system, we have 

12-10x0.707(0.3 + 1) ,_ ^, . ,, ., 
p = . Q I . o ^ -(32.2) = 4.11 ft. per. sec. per. sec, 

and hence from (4), 

tension = .^ = 12 fl - |^] =r 10.47 lbs. ; 

while s, = \pt^ = i X 4.11 X (2)' = 8.22 feet = distance de- 
scribed in the first two seconds (the velocity at end of w^hich 
= 'y^ = -pt^ — 8.22 ft. per sec). 

It is seen tliat the weight B sinks with about one eighth the 
acceleration of a free fall. 

64. Example 12. Free Fall. — A stone, allowed to descend 
freely and vertically, from rest, occupies -y- of a second (17 
watch-ticks, say) in falling through the height of a cliii ; required 

this height. From eq. (2), p. 51, M. of E. , w^e have 5 = ^^ + ^ • 

2 

In the present case c = 0, and hence the height required = -|- 
32.2 X ix(-V-)' = 290.8 ft. 

On account of atmospheric resistance, which is neglected in 
the theory of p. 51, M. of E., and is variable (being nearly pro- 
portional to the square of the velocity for the same body), the 
actual height is smaller, the discrepancy depending on the shape, 
the specific gravity, and absolute size of the falling body. If the 
stone is round, and about one inch in diameter, the average 
resistance in the above case might be as much as one-quarter of 
its weight, so that we might write \g instead of g for a rough 
approximation. (See p. 822, M. of E.) If it were two inches in 
diameter (same substance), its weight would be increased eight- 
fold, and the average resistance about quadrupled, and thus the 
latter might be about one eighth of the weight. 

"With smaller heights of fall the resistance is much smaller, 
not only absolutely but proportionally, on account of the smaller 
average velocity. 



NUMERICAL EXAMPLES— DYNAMICS OF MATERIAL POINT. 67 

65. Example 13. Block on Circular Guide (Fig. 80).— The 

A c AD smooth- curved guide ABD is smooth and fixed, 

^i^G ^71 DE rough. Qf ^he fonn of the quadrant of a circle 

^\ ,/^" I with a horizontal tangent at D. The 

^^ Id e plane DE is rough.'' The block G 

^^^^ ^^^////Wy^ \ weighs 20 lbs. and is to slide from rest 
Fig. 80. at A dowu the circular guide. How 

far (i.e., distance s^ = ?) will it slide on the rough plane DE 
before being brought to rest, if the latter offers a frictional re- 
sistance of 20 oz. (i.e., li lb.) ? The radius of the curve in which 
the centre of G moves is 48 inches. 

The velocity v^ of G on its arrival at D is the same as if it 
fell freely through the corresponding vertical height CD = 48 
in. = 4 ft. (the time of descent, however, is quite different) ; for 
the guide is both fixed and smooth (see p. 83 and also foot of 
p. 80, M. of E.). Adopt the foot, lb., and second. 

.-. V, = V2 X 32.2 X 4 = 16.05 ft. per sec. For the motion 
on DE, v^ is the initial velocity, and the motion is uniformly re- 
tarded (i.e., the acceleration is constant and negative) if we take 
the direction from D toward ^as positive ; since the only force- 
component along the path is — 1.25 lbs., the gravity-force of 20 
lbs. being -\ to the path. The acceleration isj? = force -^ mass, 
the mass being = 20 -^ 32.2 = 20 x 0.0310 = 0.620 ; .'. p = 
( — 1.25) ^ .62 = — 2.012 ft. per sec. per sec, and from eq. (3) 
of p. 54, M. of E. (in which, for present purposes, we put s =^ s^, 
V — 0, c = v^, and p as above), we have 

s, = l- (16.05)^] -^ [2(- 2.012)] = 64 ft. 

If we inquire the pressure D between the block and curved 
guide just before reaching D, we note that that pressure must 
not only support the weight of the body, but must also provide a 

proper deviating force — -— to retain it on the curve ; 

whence P = 20 lbs. + ^^i^^i^^j ibs. = 60 lbs. 
(See p. 83, M. of E.) The pressure on DE is only 20 lbs. 



68 NOTES AND EXAMPLES IN MECHANICS. 

Again, suppose the block to start from rest at B^ the angle 
BCD being 45° ; find v^ and s^ . (The acceleration on DE is the 
same as before.) 



Fig. 81. 



V, = ^2 X 32.2 X 4(1 - cos 45°) = 8.68 ft. per sec. 

s, = (0^ - V,') -r- [2(- 2.012)] = 18.75 ft. 

66. Example 14. Harmonic Motion of a Piston (Fig. 81). — 
,. On account of the great mass, and large 

radius, of the rim of a fly-wheel on the 
"\=* same shaft, the rotation of the crank is 
fr-5JD practically uniform; at least during any 
y one turn, i.e., the crank-pin is considered 
to move with a uniform velocity in a cir- 
cle. From the design of the piston and 
slot this body oscillates v^'ith harmonic motion in a horizontal 
path (see foot of p. 59, M. of E.) ; the left to right stroke alone 
is to be considered. The pressure called P is the total effective 
steam-pressure, i.e., the difference between the total pressure of 
the steam, now on the left of the piston, and the total atmos- 
pheric pressure on the right face. Friction on the guiding sur- 
faces is neglected, and since the motion is horizontal the weight 
of the piston has no component along its path. 

If P is constant throughout the whole stroke (left to right), 
and = 6000 lbs., and the crank turns uniformly at the rate of 
{u ==) 200 revolutions per minute, r being r= 8 inches; what 
must be the value of the pressure P' between crank-pin and the 
side of the slot just after the dead-point C is passed, i.e., at the 
beginning of the stroke? Ditto^ when the crank-pin is 45° from 
C\ and again, when it is at 0^ 90° from G% The weight of the 
piston and rod is 160 lbs. 

Between C and 0^ P' is smaller than P and is a resistance, as 
regards the motion of the piston. If that motion were uniform 
the full amount of the 6000 lbs. would be felt at the pin, for in 
that case the acceleration would be zero and the horizontal forces, 
P and P' ^ would be equal and oppositely directed ; but from G 
to the motion is accelerated (the piston has no velocity at C\ 



HAEMONIC MOTION OF PISTON. 69 

SO that a certain amount, = mass X accel., of the 6000 lbs. is 
absorbed in the " inertia" of the mass, so to speak, leaving only 
the remainder to be felt as a pressure P' at the crank-pin. This 
is expressed analytically by the relation 2 (hor. comps.) = Mp ; 
i.e., F- P' = Mp. 

Beyond O toward D the constraint of the mechanism is such 
as to bring about the gradual stopping of the piston, which at 
has its greatest velocity (= c, = to that which the pin has at all 
times), so that independently of the 6000 lbs. on the left the piston 
is, as it were, thrown against the crank-pin, the pressure produced 
against which at any instant from to D must = Mp over and 
above the 6000 lbs. due to steam action ; i.e., P' = 6000 lbs. 
-\- Mp (where J? is the numerical value of the acceleration, whose 
algebraic value is now negative), or, analytically, P — P' = Mp 
where p has its algebraic value (negative when a number is in- 
serted for it). 

The linear velocity of the crank-pin is = '^nrii, = 2 X -^^ • | • -W'? 
= 13.97 ft. ]3er sec. (using the ft., lb., and sec). As the pin ap- 
proaches and passes a dead-point the motion of the foot of the 
perpendicular let fall from it upon the horizontal diameter, along 
that diameter, is not only the motion of the piston, but is at this 
point normal to the path of the pin ; hence the normal accelera- 
tion of the pin is the actual acceleration of the piston at a dead- 
point. Therefore & -^ r (see eq. (4), p. 75 ; and also section § 75, 
and first line of p. 60, M. of E.) is the acceleration of the piston 
at C\ and therefore just after passing the dead-point C we 
have 

r ozi f 

= 6000 - 1452 = 4548 lbs. 

The acceleration of the piston is proportional to the displace- 
ment, and hence at 45° from O we have 



r— „2 



- cos 45°1 = 6000 - 1452 X .707 = 4937 lbs. 
r J 



P' = P-M 

At 0,p = 2ind P = P' = 6000 



70 NOTES AND EXAMPLES IN MECHANICS. 

Bejond at 45° 

P' = 6000 + 1452 X .707 = 7063 lbs. ; 
while just before reaching D 

P' = 6000 + 1452 = 7452 lbs. 
On the return-stroke steam is admitted to the right of the piston 
and P^ occurs on the other side of the slot, with same variation 
during the stroke as before. 

67. Example 15. Conical Pendulum, or Simple Governor-ball 
(Fig. 82 [a]). — If the oblique part of the cord is to be 20 in. in 

length, what tangential velocity in 
...^ /"a--! a horizontal circle (centre at C) and 
/f I what radius = r, for that circle, must 
"i be given to the material point G of 
10 lbs. weight {= G) in order that 
motion in the circle shall be self- 
[a] |T|G=ioz6s. [bj perpetuating and the weight G' of 
Fig- 82- 40 lbs. may be sustained at rest 1 

Fig. 82 P] shows the moving weight as a free body, the only 
forces acting being a gravity-force of 10 lbs. and an oblique cord- 
tension, P, which by above conditions is to be 40 lbs. The mo- 
tion of G being confined to a horizontal plane, it has no vertical 
acceleration ; therefore 2 (vert, compons.) should balance, or 

P cos <af — 6^ =: 0, whence cos « r= -^ — 0.25, and a should 

= 75° S""'. Hence r should be made = (20 in.) x sin a = 20 

X 0.968 — 19.36 in. == 1.613 ft. Since the motion is to be in a 

curve, '2 (no'^mal compons.) at any instant should = J!/ X {vel.y 

G & 
-7- rad, ; i.e., P mv a-\-^ ^^ — - — \ and combining this with the 

P COS arzz G derived above we have tan a=: c^ -r- gr^ whence 
the required velocity must be c = 4/32.2 X 1.613 X 3.871 = 14.2 
ft. per sec. The proper radius being as above {r = 1.613 ft.), 
this implies rotation about C at the rate of 

u = - — = - — -- ' ^^^ = 1.40 revolutions per second, * 
27rr 271 X 1.613 ^ 

or 84 per minute. 



CONICAL PENDULUM — EXAMPLE. 71 

68. Example 16. The Weighted Governor or Conical Pendulum 
(Fig. 83). — The four inextensible cords, each 16 inches long, 
connect the three "material points," or small ,/////,/,,,^A y/M,,,,,,/7. 
bodies, as shown ; the two upper cords being at- //^\ 
tached to a fixed support at J.. G, and G^ are //§' \-o^\~^^ 
two balls of equal weight; the weight of each ^-::il!ll^":::.^ 
= 6^1 = 8 lbs.,' while the block G^ weighs 12 ^^"1"^^^ 
lbs. If now the balls are caused to rotate about ^S./^^ 
the vertical axis ^ 6^^ at slowly increasing rate g_8 76^jS'° 2=^2^6?. 
(revolutions per minute), by pressing against them f^^- ^3. 

laterally with a vertical board whose plane contains the axis 
A GB^ the angle a gradually increases and the block G^ is lifted 
along the axis towards A. When the speed of rotation has 
reached any desired figure (rev. per min.) a has some correspond- 
ing value, and if the board is now removed a retains that value 
and the balls continue their motion (forever, if no friction) in the 
corresponding horizontal circle, sustaining the block G^ at rest 
(at least its centre of mass is at rest) in some position B. 

Required the distance AB^ = 2^(7, when a speed of rotation 
of 120 revs, per min. has been attained \ 

Let I denote the cord-length of 16 in., r the unknown radius 
A of the horizontal circle, S^ the tension induced in 
each of the upper cords, S^ that in the lower, and c 
the unknown linear velocity of each ball. Let u = 
the number of revolutions per unit time (so that 
c = ^Ttru). Fig. 84 shows one of the balls as a free 
body. Since its vertical velocity is always the same 
(zero), i.e., its vertical acceleration = (the motion 
being confined to a horizontal plane), 

^ (vert, comps.) = ; i.e., S^ cos a — S^ cos a — 6^, = ; . (1) 

while on account of the curvilinear motion ^ (normal compons.) 
= Mg^ -t- r, or 

G & 
/S', sin a ~|- /Sj sin a = — ^ • - (2) 





G2 



72 NOTES AND EXAMPLES IN MECHANICS. 

The tangent to the curve is -| to the paper and ^ (tang, comps.) is 
evidently = 0, whence the tangential acceleration 
mnst be zero ; i.e., c is constant, as we have as- 
sumed all along. 

With G^ free, in Fig. 85, we have balanced 
forces; whence 2 (vert, compons.) = 0; i.e., fig. 85. 

2x^^ cos ^ - 6^, = (3) 

By elimination, noting that c = 27rru, 

c^ (6^,+ Qtan^ ^ _r^ _ ^g_ G,-^G, 

gr~ G, ' ' tan «' - ^^' " 47rV * G, ' 

We note, therefore, that the required distance AC \^ inde- 
pendent of the length I and is inversely jprojportional to the square 
of the number of revolutions per unit time. (A similar result 
was found with the simple conical pendulum ; see p. 78, M. 
of E.) 

Hence, numerically, with the foot, pound, and second (so that 
u = -i-g^/ = 2 revs, per sec), 

2x32.2 8 + 12 




AB, - 2AC\ = ^ ^ ^3^ ^ ^ • -^ = 1.018 ft. ; or 12.22 in. 

69. Example 17. Cannon-ball under Gravity and Air-resistance. 

— A round cannon-ball, weighing 24: 

lbs., is at a certain point of its path 

moving with a velocity of v = 800 ft. 

per second in a direction making an 

angle of 20° below the horizontal. The 

G=2Ubs. resistance offered to it by the air at this 

F'<^*s^- speed is 80 lbs. and acts in the line of 

motion, since the body is round and has no motion of rotation. 

Required the amount and position of the (ideal) resultant force 

B, See Fig. 86. 

Since there are only two forces acting on the ball, P and G, 
R must have an amount and position determined by the diagonal 
of the parallelogram formed on P and G. See figure for the 
known angles. Plence (from formula on p. 7, M. of E.) 
R = VP' +G' + 2PG cos 110°, i.e., 



P = VSO' + 24'^ + 2 X 80 X 24 X (- 0.3420) = 75.25 lbs. 



J^^UMERICAL EXAMPLES — DYNAMICS OF MATERIAL POINT. 73 

To find the angle 6^ note that in the triangle PRO we have 

24 

sin 6/ : sin 70° :: (9^ : ^; whence sin d = =^-^(0.9397); i.e., 6 = 

17° 26', and hence Id is 2° 34' above the horizontal. 

Since the action-line of R is not coincident with the line of 
motion of the ball at this instant, the path of the hall must he 
curved^ the radius of curvature at this point depending on the 
mass, on the square of the velocity, and on the value of the 
normal component of JR ; while the rate of retardation of the 
velocity (negative tangential acceleration) depends on the mass 
and the tangential component of R. (See next example.) 

70. Example 18. Ball in Curved Path. Radius of Curvature, 
etc. (Fig. 87). — A large ball weighing 200 
lbs. (so that its mass = J/ = 200 -^ 32.2 
z=z 200 X 0.031 = 6.2, in the foot-pound- 
second system of units) at a certain point 
of its path has a velocity of 700 ft. per 
sec, the resultant force R at this instant 
being = 300 lbs. and making an angle of 
140° with the direction (see v in figure) of motion. 

Required the radius of curvature, r, at this point of the path, 
and also the tangential acceleration. 

By a rectangular parallelogram of forces we resolve R along 
the tangent and normal, obtaining for its tang, compon., 

T,=R cos 140°, =r 300 x (- 0.76604) = - 229.812 lbs., 
while ]Sr,=R sin 140°, = 300 x 0.6428 = 192.84 lbs. 

From eq. (5), p. 76, M. of E., ^ (norm, comps.) = Mv" ~- r^ 
and 2 (tang, comps.) = Mp^ , whence 

6.2x(700r_ 

192.84 -^^W^^it., 

- 229.812 ^ ^ 

and pt = ^"o = — 37.06 ft. per sec. per sec. 

This last value means that the retarding effect of the com- 
ponent Tis such that ^the rate of retardation remained constant 




74 



NOTES AND EXAMPLES IN MECHANICS. 



^-n Gi 



l5-A+-G-> 

" r""Hi""iR 



for one second, at the close of that second the velocity in the 
path would be 700 - 37.06 = 662.94 ft. per sec. 

71. Example 19. Steam Working Expansively and Raising a 

Weight.— In Prob. 4 of p. 61, M. of 
E., supposing the boiler-gauge to read 
80 lbs. per sq. in. (above one atmos- 
phere) and the total length of stroke, 
Sn = OJV, to be 16 inches, with cut- 
off at one third stroke (so that s^ = i 
of 16 in.), the diameter of piston 
being 10 inches ; how great a weight 
G can be raised if the (circular) pis- 
ton is to come to rest at the end of the stroke, having started 
from rest at the beginning of the stroke? Required also the 
time occupied from to £, and the position of the piston when 
its velocity is a maximum. 




Fig. 87a. 



From p. 62 we liave the equation 



<S...[l + l0g,(j)]=^5„+^.„, 



(2) 



now to be solved for G. 5^ = f ft. ; ^^ = |^ ft.^ and hence the 
ratio Sn : s, = 3. The area of piston = nr" = ^- x (5)' = 78.57 
sq. in. 

.'. Air-pressure above piston, = ^, = constant = 78.57 X (15 
lbs. per sq. in.) = 1178 lbs. ; while the steam-pressure under pis- 
ton while it is passing from io B^ =. S^^ — 78.57 X 95 = 
7464.15 lbs. Noting that loge = common log X 2.302, we have 
from eq. (2) (using the foot, pound, and second) 

7464 X i[l + 2.302 X .47712] =r 1178 x J + 6^ X f 

Solving, G = 4044 lbs. (so that M = 4044 x .031 = 125.36). 

The acceleration from 6^ to jS is constant and = 
^^ ^ ^S^ -A-G)-'^M= 2241 -^ 125.36 = 17.88 ft. per sec. per 
sec. ; and [eq. (2), p. 54, M. of E.] 5, = \pf^ ; hence 



time from Oto B = t, = \/ jy^ = V.0496 = 0. 



222 sec. 



BALL FALLING ON SPRING. 



75 



Above B, the steam-pressure S diminishes, and when at some 
point m it has become = A -\- G, i.e., to 5222 lbs., the resultant 
or accelerating force, S —{A -\- G), is zero ; above this point m 
that force is negative, i.e., the velocity diminishes, and hence the 
velocity is a maximum at m. Let s^^ be the distance of m from 
O ; then from Boyle's Law s, : 6^^: : 5222 lbs. : S^ , whence 
s^ = l-ltfl- = 0.635 feet, = 7.620 in. 

72. Example 20. Ball Falling on Spring (Fig. 
weighing two pounds {G) falls freely from rest, 
and after falling 5 ft. {-— h) comes in contact with 
the head of a spring, which it gradually compresses 
during its further descent until brought to rest 
again momentarily (at m^). The resistance {P) of 
the spring is proportional to the depth of compres- 
sion {s) and is 60 lbs. {P^ at the end of the first 
inch (s^, (Provision is made against side-buck- 
ling.) Required the maximum compression onjn^^ 
= s,. 

At the end of the free fall the velocity of the 
balls is c = V2gh (i.e., c^ = 2^^^), since so far there is 
but one force acting, its own weight G. At any instant., 

distance 5, however, below m. (the point of first 

contact) the resultant downward force is 6^ — P^ fig. 88. 

P being the upward pressure of the head of the spring against 
the ball at this instant, and the acceleration is therefore variable 
and is j9 = force -^ mass, — {G — P) -^ {G -^ g). Let down be 
positive. Substituting in vdv — 2^ds, noting that P:P^::s:s^^ 
and then integrating between the points m^ and m, , we have 




^ 



-vdv = ds 



,i[»-£l 



1 1 r r^ p C' 

-pPds ; and — / vdv = I ds — ^r I ^^^' 



c 
Numerically, with the inch, pound, and second, 

60 s; 



— 60 = 5, - 



2 X 



— ; whence 5/ — ~s^ 



Finally, m,m, , = s,, = 



1 2 ' 

2.03 inches (and 



15 
1.96 in.). 



76 NOTES AND EXAMPLES IN MECHANICS. 

The negative result refers to a point (call it m') 1.96 in. above 
m^ . This is the position where the ball would momentarily come 
to rest for the second time, if it adhered to the head of the spring 
after the latter had regained its natural length, supposing the 
lower end of the spring to be fixed. This motion of the ball 
while in contact with the spring is really harmonic^ whose central 
point, from which the "displacement" would be reckoned, is 3^ 
of an inch below m^ , i.e., at the point where the pressure of the 
spring =: 2 lbs. (the weight of ball), so that as the ball passes that 
point its acceleration is zero and the velocity a maximum. This 
point is midway between m^ and m! , 

72a. The Engineer's " Mass." — The engineer measures the mass 
of a body (in case a problem connected with its motion is under 
treatment) by the fraction, weight — accel. of gravity ; or G-^ g. 
This is not scientific, but is so firmly rooted in engineering prac- 
tice that no different measure can well supplant it. It seems 
to imply that the amount of matter in a body depends on the 
existence of the attraction of gravitation ; whereas, of course, 
such is not the case. This measure arises from the fact that a 
convenient way for the engineer to determine the magnitude of 
any force I^ (or resultant) acting on a body and producing an 
acceleration (j?) of its velocity is to compare it with the force of 
gravity exerted on the body, whether the circumstances of the 
problem are affected by gravitation or not. In the phrase force 
= 7i%ass X uccel.^ or I^ = Mp, the word mass is simply a name 
given to the fraction G -^ g, the origin of which is as follows : 

In the actual problem the force I^ produces an acceleration 
= J) in the velocity of the body. In the ideal experiment of 
allowing the same body to have a free fall m vacuo we know 
that the only force would he the weight G, and that the resulting 
acceleration would he g ; and since the forces must be propor- 
tional to the accelerations, we have (§ 54, M. of E.) 

G 
P : G :: actual p : ideal g\ or, P — — ^. 

tj 

In other words, the engineer uses the gravitation measure of a 

force (p. 48, M. of E.). 



CHAPTER y. 



Moment of Inertia of Plane Figures. 

72b. Phraseology. — Unless otherwise specified, we are to under- 
stand by " moment of inertia of a plane figure" the rectangular 
moment of inertia ; i.e., the axis of reference lies in the plane of 
the figure (and not ~[ to it as with the " polar" moment of in- 
ertia). This is a useful function of the plane figure, to be used 
in the theory of beams under bending strain. 

73. Moment of Inertia of Section of I-beam (Corners not Rounded). 
— Ficr. 89 shows ttie form and dimensions of fl 

• • r~^ ^^ 1 ^ 

the section, which is symmetrical about each T j[^4^^?^!" 
of the axes X and Y, and is for present pur- i , 
poses subdivided into three rectangles and f I 
four right triangles. Making use, then, of , ,, 
results obtained for those elementary forms, -^^ 
and of the transferral formula between the 
gravity axis of any figure and a parallel axis 
(see p. 94 and eq. (4), p. 93, of M. of E.), 
we have for the moment of inertia about 
axis X 



'W 



-\-la .d' 



h'h' 



[■ 



36 



\12 ' —- j 1 12 
That is, numerically, 
2[TV-(6)-(|y + 6-|-(8|iy] + TV4-(16i)' +4[3V 




•(1)' 

+ V-(8.04)'] 
= 566 + 196.8 + 356 = 1117.8 bi. in. = I, . 
Similarly, the moment of inertia about the axis Y is 



2- 



aP . h'h' 



+ 4U, 



U-A 



2 ' 



{I - V) 



• d"' 



12 ' 12 ' ^L36"^ \2 2/ ' 2 2 

= ^'^-m' + (leDrVtt)^ + 4[3V(i)(-yT + *-v-(i)ar] 

= 22.5 + 0.17 + 9.79 = 32.46 bi-quad. in. = I^ . 

77 



78 



NOTES AND EXAMPLES IN MECHANICS. 




All plates h thick 
Fig. 90. 



74. Moment of Inertia of a Section of a Built Box-beam (Fig. 90). 

— The beam is composed of two 
'' flange-jplates^'^ (upper and lower), 
two vertical '^ stem-plates ^^^ and 
four ''^ angle-hars^^ of equal legs, 
riveted together. See figure for 
notation and dimensions. Ke- 
quired the moment of inertia 
of the whole section about X, its 
horizontal gravity axis of sym- 
metry. 

In Fig. 91 we have the loca- 
tion of the gravity axis g (parallel to X) of a single " angle" 
section, according to the hand-book of the New 
Jersey Steel and Iron Co., so that from the dis- 
tance 1.68 in. we compute the d' = 10.32 in. of 
Fig. 90, or distance of axis g from axis X. 
From the same book we find that the Ig of the 
angle-section is 20 bi. in. (very nearly), and its 
area F' = 5.75 sq. in. Let t = thickness of all plates = |- in. ; 
and t' = diam. of rivet-holes == f in. 

First, neglecting the rectangular gaps made by the rivet-holes, 
we have the 7^'s of the various component sections as follows : 

(four " angles") . . . 4[7^ + 7"^(i'^)]=:4[20+5f (10.32)^] =2528 ; 
(flange-plates) 2[-^\hf+Ud')]=2m-{iy+^^-{12iy]=S001; 

(stem-plates) ^[j^'] =^[^^1 =1152; 

making a total of 6681 bi. in. 

Treating the small rectangles left by the rivet-holes as con- 
centrated in their respective centres of gravity [and thus neglect- 
ing their local (gravity) moments of inertia], 

Subtract! ve J ._ //)'\2-i 

/, due to t =4:[{2t)'fd'''']+^]ji2ty[-^j J = 243 + 432=675. 
rivet-holes ) 

Hence, ^ of actual section = 6681 — 675 = 6006 bi. in. 



I il.68" 

^ Section of 
Y/ Angle-Bar 
I 6x6x^" 

Fig. 91. 



MOMENT OF INERTIA BY SIMPSON'S RULE. 



79 




It will be noticed that the first term, -j^M [or local (gravity) 

moment of inertia], in the I^ of the section of a Jlange-plate, 

above, is very small compared with the second, or " transferral 

term" {ht , d''). This is due to the fact that all parts of a thin 

flange-plate section are very nearly at the same distance from the 

final axis of reference, X, In such a case it is customary to 

neglect the local term, as no practical error results from so 

doing. 

75. Moment of Inertia of Irregular Curve-bounded Plane Figures 

by Simpson's Rule (see § 93, M. of E.). — If an exact result for the 
7^ of the figure shown in Fig. 92 were desired, 
we might first conceive of its subdivision into 
narrow strips parallel to X^ of variable length 
V and infinitesimal width dz^ then express its 1^ 
as y (small area) X ^^ ovf{vdz)z'^^ =^f{z'^v)dz ; 
and finally perform the integration, if v were 

an algebraic function of z. If such is not the case, however, 

(or if such is the case but the in- 
tegration net practicable,) we can 

resort to Simpson's Rule (§ 15), 

noting that the u^ x, and dx of 

that rule correspond to the {z^v\ 

0, and dz^ respectively, of the 

present problem. We divide the 

whole height of the figure (from 

the axis X) into an even number 

n of equal parts, and through each 

point of division draw a parallel 

to X, thus determining a series 

of widths, 'Wo, -y, , '^2? ^tc. For 

example, this construction being 

made for the upper part of the rail-section in Fig. 93, withn 

we have its /,,, =f{z''v)dz^ approximately = 




Fig. 93. 



6, 



3X6 



{z:v, + 4.{zx + ^>, + K'v:) -f 2(.,x -f z:v,) + ^,vj. 



With numerical substitution, therefore, the lengths marked in 
the figure having been scaled in fiftieths of an inch (noting that 



80 NOTES AND EXAMPLES IN MECHANICS. 

z, = 0, z,= ^B^ , 2^ — p, , 2, = p, , etc.), we have, as the I^ of 
the portion of the rail-section lying above the axis X, 

^^JO + 4(12 X 40 4- 32 X 80 + 5-^ X 163) -f 2(2^ x 50 + 4^ X 136) + 6'^ X 141] 

bi-quad. fiftieths ; which divided by (50)*, or 6,250,000, gives 

I^ for upper part = 25.27 bi. in. 

Similarly, the vertical height of the lower part being divided 
into four equal lengths, we have for the I^ of that part (nearly) 

|^|^3 [0 + 4(1^ X 53 + 3^ X 2M) + 2(2^ X 103) + 4^ X 266] 

= 129,870,000 bi. fiftieths. Dividing by (50)*, we have 20.77 
bi. in. ; so that the total Ix of the complete rail-section = 20.77 
+ 25.27 = 46.04 bi. in. 

JTis a gravity axis parallel to the base of the rail-section, and 
has been located by cutting out the shape from card-board and 
balancing on a needle-point. 

If the plane figure is of such a form that a division into strips 
perpendicular to, and all terminating in, the axis 
of reference (JT) is convenient, the exact cal- 
culus form for its Ix is \fifdx (see latter part of 
§ 93, M. of E.), for each strip is an elementary 
rectangle. See Fig. 94. 
"^g " ' If Simpson's Rule is to be applied, divide the 
Fig. 94. base OX of the figure into an even number, n^ of 

equal parts and scale the extreme ordinates y^— AO 2XiA y^— BX^ 
also the intermediate ordinates y^ , y^ , etc., at the points of divi- 
sion. "With 71 r=: 6, for example, we have as an approximation 
7TX 

I, = ^^[2/0^ + %.^ + y.' + y:) + %; + y:) + y:]' 

76. Graphical Method* for the Gravity Axis and Moment of 
Inertia of a Plane Figure (Fig. 96).—A''B''0'' is the figure in 
question (drawn in full size). It is required to construct the 
special gravity axis, Ji, .that is parallel to the base A'^B'', and 
to obtain the moment of inertia about that axis. 

Divide the figure into strips parallel to A''B", of small width 
* From Ott's Graphical Statics. 



MOMENT OF INERTIA — GRAPHICAL METHOD. 



81 



(no widtli being more than one eighth, saj, of the total width -[ to 
A" B"^. These widths need not be equal. 

Through the centres of gravity of the strips (1, 2, 3, etc.) draw 
indefinite lines (1 . . 1', 2 . . 2', etc.) |1 to A" B" (with most strips it 
is accurate enough to take the centre of gravity midway between 
the sides). Along any right line 0' I^ parallel to A"B'\ lay off the 
lengths O'A^ AB, BO, etc., proportional, respectively, to the areas 




Moment of 

Inertia 

Graphic Method. 



Fig. 95. 

F^^F^, etc., of the successive strips, m the order and position 
show^n (in most Instances * each such area may be assumed pro- 
portional to the length of the strip measured through the centre 
of gravity). Through 0' and / draw lines at 45° with 0' I, as 
shown, to determine the "pole" 0, from which the *' rays" OA, 
OB, etc., are now drawn. Then through any convenient point 
Z draw ZO'" \\ to 00'. From the intersection ^ with 1 . . 1' 
draw cib \\ to OA to intersect 2 . . 2' in some point 5, then ho \\ 
to OB. and so on ; until finally, through i, iO is drawn || to 10 
to determine by intersecting ZC". The required gravity axis 
E passes through C \\ to A"B'\ (For proof consult § 376, M. 
of E.) 

I'he moment of inertia about axis R may be obtained by 

omdtiplying together the area of the given figure A" B" 0" (call 

it F) Ijy the area, (call it F'^ of the " inertiafigure''' (i.e., the area 

included between the two lines Ca and Ci, and the broken line, 

* When the strips are narrow and of equal width. 



82 NOTES AND EXAMPLES IN MECHANICS. 

or '* equilibrium-polygon," dbcdefghi\ shaded in Fig. 95). The 
proof of this relation (strictly true only for infinitely narrow 
strips) is as follows : 

At any vertex of the equilibrium-polygon, as at ^, there are 
two segments meeting; prolong them to intersect the gravity 
axis R in some two points, as m and n. Then nnng is a triangle 
with base m . .n (call it Ic^) and altitude a?, . But on the left of 
Fig. 95 the shaded triangle OF^ is evidently similar to mng) 
whence the proportion Jc, \ x^ : : F^ \ ^F; i.e., i^,cc, = \F\ . 
Multiplying by a?, , we have F^x^' = F{ix,Jc,). 

]^ow F^x^^ is the moment of inertia (about Ic) of strip 
"No. 7 (considered infinitely narrow), and ^x^Jc^ is the area of the 
triangle 7n7ig. We have therefore proved that the moment of 
inertia of any one strip is equal to the product of the whole area 
F of the given plane figure by the area of a triangle like mng 
and obtained in a similar manner. If all the triangles like mng 
were drawn, their united areas would evidently be that of 
the " inertia-figure^^ dbcdefghi- C-a. Hence the sum of the 
moments of inertia of all the strips, i.e., the moment of inertia 
of the whole figure A"B"G", is 

J _ \ cirea of jplane \ ^ \ ^^^<^ ^f ^^^ \ 
^ ~ [figure A" B"C" \ ^ \ ''inertia-figure'' \ ' 

In practice these areas are most conveniently and accurately 
obtained by means of a planimeter; otherwise, subdivision into 
small trapezoidal strips may be resorted to. If the scale of the 
drawing is one half of the actual size the result must be multi- 
plied by 16, i.e., the fiourth power of 2 ; and similarly for other 
ratios. 

By the use of a planimeter with this actual figure (Fig. 95) 
the results F= .90 sq. in., and F' = .42 sq. in., have been ob- 
tained ; therefore Ir = 0.378 bi. in. 



CHAPTER YI. 



Dynamics of a Rigid Body. 




Fig. 96. 



77. Example of Rotary Motion. Axis Fixed and Horizontal 

(Fig. 96). — The body AB consists of an irregular solid and a 
light drum, rigidly connected and mounted on 
a horizontal axle whose journals are 2 inches 
(= ^r) in diameter, at C \ the radius of the drum 
being <a^ = 5 in. The weight of AB is 150 lbs. 
= G^ , and its centre of gravity is situated in the 
axis of rotation (hence G^ has no effect as regards 
changing rotary velocity, having always a zero 
moment about the axis). The weight (x, = 20 

lbs., is attached to an inextensible cord which is wound on the 

drum and whose weight is neglected. A constant friction i^ 

= 10.5 lbs., acts at the circumference of the journals. 

It is required to compute the radius of gyration he of the 

rotating body by the experiment of noting the time t^ occupied 

by G in sinking a measured distance 5, , from rest. Suppose 

^j = 5 sec. and s^ = 10 ft. 

At any instant during this motion, the tension in the cord 

being S, we have for the angular acceleration 6 

of AB, which is shown free in Fig. 97, taking 

moments about axis O, 






Fr, 




Fig. 97. 



while at the same instant the downward linear 
acceleration j9, = 6a, of G^ enters into the rela- 
tion involving the sum of downward forces for 6^ as a free body, 

viz.; 



O ^ S = mass. X accel, = —p, 

9 



(2) 



84 KOTES AND EXAMPLES IN MECHANICS. 

Solving forp, we would find it constant, as also /S; hence G 
has a uniformly accelerated rectilinear motion, and AB a uni- 
formly accelerated angular motion. Therefore (with foot, pound, 
and second), since (from eq. (2), p. 54, M. of E.) 5, = ^pf, and 
hence ^:> = 2,9^ -^ t^"", we have j9 = 20 -^ 25 == 0.80 and is the con- 
stant linear acceleration, not only of G, but for any point in the 
surface of the drum, so that the angular acceleration of body A£ 
is = p -^ a = O.SO -^ YJ — 1-^^ radians per sec. per sec. From 
eq. (2) we now have S = 20 — (20 X .80) -^ 32.2 = 19.504 lbs. 
as the constant tension in the cord ; and finally, from eq. (1), 

. _ [19.504 X A - 10.5 X A]32.2 _ , 

^' - 1.92X150 -0.blJsq.it., 

i.e., the rad. of gyration ^^ = 0.901 ft. = about 10.8 inches. 

78. Solution of Example of Compound Pendulum. (For the 

statement of the example see p. 121, M. of E.) Fig. 98. (A =6'', 
7' = 1.2". Locate axis oo so that the time of 
oscillation t^ shall be J sec.) Referred to the 
horizontal-gravity axis cc we have for the solid 
cone (from § 101, M. of E.) 7, = /^i¥[r + M^], 
i.e., K = Mr' + m = A[l-4^ + ¥-] = 1.566 
sq. in. With the inch and second g = 386.4, and 
Fig. 98. gt'^ -^r^Tt' = 4.889. Hcncc, from eq. (1), p. 120, 

M. of. E., 




^ ^ 4 889 + 1^23.90 — 1.556 = + 9.615, and + 0.163 in. 

That is, with either oo or o'o' as axis of suspension, co being 
made = 0.163 in., and co' — 9.615 in., the duration of an oscilla- 
tion (of small amplitude) will be \ sec. of time. As a check we 
note that co + co' = 9.778 in., which = 2 X 4.889 = length of 
a simple pendulum beating half-seconds (see above and also eq. 
(3), p. 119, M. of E.). 

79. Points of Maximum and Minimum Angular Velocities Id 
Motion of Crank-pin of a Steam-engine. — Fig. 99 shows the results 
of applying the tentative graphic method mentioned in the mid- 
dle of p. 124, M. of E. In making trials for successive positions 
of the connecting-rod, advantage can be taken of the fact that if 



EOTATION — EFFECT ON BEARINGS. 



85 



the point be found where the axis of the connecting-rod (pro. 
longed if necessary) intei sects the vertical line (vertical in this 
instance), drawn through the centre C of the crank circle, the 
distance of that point from C represents the amount of the 
tangential component, T^ of the force P' ^ on the same scale on 
which the length, C . .n^ of the crank would represent the force 
P, Hence various positions of the connecting-rod are drawn 










P=550076«.a4«M)ne* 
const.during stroke. 

Connecting-Eodji ^ 
beiween.cenlres; 




%" 



Circle 



Fig. 



until those two are found for which the distance mentioned above 
is fl, i.e., ^, of radius G . . n. 

In the working of most engines the steam is used expansively/ 
so that P is variable, being greatest near the beginning of the 
stroke. This would cause the points ?i and m to be nearer to A ; 
and similarly, on the return-stroke, n^' and m'^ nearer to £. 

80. Rapid Rotation of a Body on a Fixed Axis. Effect on 
Bearings. — So long as a wheel or pulley in a machine is perfectly 
symmetrical about the axis of rotation the pressures on the bear- 
ings are due simply to the weight of the pulley and the pulls or 
thrusts of cogs, belts, cams, etc., which may be acting on the 
pulley or on the shaft on which it is mounted ; whether rotation 
is proceeding or not. But if, through any imperfection in the 
adjustment or mounting, the centre of gravity lies outside of the 
axis of rotation (its distance from that axis being p), other press- 
ures are brought upon the bearings, the'Bame in amount (2^ ^the 
pulley were at rest and a force = 00" Mp acted through the centre 
of gravity, away from, and "i to, the axis of rotation. 



86 



NOTES AND EXAMPLES IN MECHANICS. 




For example, let the pulley in Fig. 100, of weight = 644 lbs., 
be out of centre by one fourth of an inch and be 
rotating uniformly at the rate of 210 revolutions 
per minute, i.e., f rev. per second. Its angular 
velocity is therefore cb? = 22 radians per second 
and the pressures on the bearings due to the eccen- 
tricity of the pulley at this speed is the same as if 
the pulley were at rest and a " centrifugal for ce^^ P^ = [484(644 
-i- 32.2) -^y, = 201.6 lbs., acted on the body, as shown in figure, 
where G is the centre of gravity. The pressures on the two 
bearings due to this ideal " centrifugal force" will be inversely 
proportional to their distances from the pulley-centre and parallel 
to P (aside from the pressures due to the weight of the pulley, 
action of belts, cogs, etc.). 

The continual change of direction of these centrifugal press- 
ures, as the wheel revolves, is likely to cause injurious vibrations 
in the framework of the machine. 

81. " Centrifugal Couple.^^ (Continuing the discussion of the 
last paragraph.) — Even if the axis of rotation does contain the 
centre of gravity (or rather centre of mass in this connection) of 
the revolving body ; if, even then, we find that any plane 
containing the axis divides the body into two parts,* the right line 
connecting whose centres of mass is not ]?erj)endicular to the axis, 
the ideal centrifugal forces, oj^M^p^ and co'^M^p^^ of these two 
component masses form a couple, causing pressures at the two 
bearings ; which two pressures, themselves, form a couple. For 
example, Fig. 101, where the axis of the shaft AB and the cen- 
tres of the two masses are in the same 
plane, and where the products M^p^ and 
J/2P2 are equal, the centre of mass of 
the whole rigid body must lie in the 
axis of rotation (i.e., neglecting the mass ^^ 
of the shaft) ; but the line joining the C^^^ 
centres of the component masses is not ^^ 
perpendicular to the axis AB. A and p' " 
B are fixed bearings and a uniform ro- 
tation with angular velocity of od is proceeding. 

* We here suppose each of these parts to be homogeueous and to be sym- 
metrical about a plane passed through its center of gravity and perpendicular 
to tile axis of rotation. If such is not the case, we must resort to the principles 
of ^§ 123-122C iuclus. of M. of E., to determine the pressures at the bearings. 



Pi 



Ml 



r^. P' 






'I 



.1^. 



" Fig. 101. 

Evidently the 



PRACTICAL I^OTES Ol^- PILE-BEIVING. 87 

action on the bearings is the same as if the bod}' AB were at 
rest and were acted on bj the two ideal centrifugal forces 
P^ = Gj^M^p^ and P^ = c^M^q^ , in the positions shown in the 
figure (on the right). These forces are equal (since M^p^ = M^p^^ 
forming therefore a couple, so that the reactions at the two bear- 
ings (pressures of bearings on shaft) would form a couple of 
moment P'h = P^a. 

For example, let J/, weigh 120 lbs. ; M^ , 100 ; the distances 
Pj and P2 being 2 and 2.4 ft., respectively ; while the angular 
velocity is 22 radians per second (210 revs, per min.). Also let 
a = S it. and & = 4 ft. ; then 

F' = fp, = l-m'~ X 2.4 = 2710 lbs. 

So long as the rotation is uniform these pressures P' lie in 
the plane of the axis and the two mass-centres ; which plane, of 
course, is continually changing position. 

82. Piles. (See p. 140, M. of E.) — As a practical matter it 
should be understood that as the head of the pile becomes broomed, 
or splintered, during the driving, the penetration occasioned by 
the blow may be much smaller than would otherwise be the case 
(only one quarter as much in some cases) ; hence it is economical 
of power that the head be adzed off occasionally, and especially 
should this be done just previously to the last few blows when 
the measurement of penetration is made, to be used in a formula 
for safe bearing load. 

A formula for the safe load has been proposed (see p. 185, 
Eng. News, Feb., 1891) in which, besides adopting the divisor 6 
of eq. (2), p. 140, M. of E., the value of s' is assumed one inch 
greater than that actually observed. In cases where the actual s' 
is small (as an inch or less) this allows a very wide margin of 
safety. If s' is large, the margin of safety is practically little 
more than that afforded by the divisor 6. This formula may be 
written : 

^ . 7 . _ 1 Gh ( or, for the ) _ 1 Gh 

baje Load - g • ^y 4, ^^^^ •^^;, . \ ^^ and ft, S ~ 6 -^^ + tV 



88 NOTES AND EXAMPLES IN MECHANICS. 

while if G is in lbs., h in feet, and s^ in inches, it takes the form 

2Gh 



safe load in lbs, = 



+ !■ 



(See Baker's Masonry Construction, and Trautwine's Pocket 
Book, for many practical details in the matter of piles.) 

In experiments in the laboratory of the College of Civil 
Engineering at Cornell University, with round oak stakes 2 in. 
in diameter and a 4-lb. hammer falling four feet, the actual bear- 
ing loads have been found to be from about one half to seven 
eighths of that given by the expression Gh -^ s\ 

83. Kinetic Energy of Rotary Motion. Numerical Example. — 
The grindstone in Fig. 102 has an initial angular velocity, gd^ , 

corresponding to 180 rev. per min., about 
its horizontal geometrical axis, which 
is fixed. How many turns will it make 
before it is brought to rest by the two 
frictions, at A and at C^. The pressure 
of the plank OD on the stone at A is 
due to the weight of 180 lbs. suspended 
at rest at D. Assume that the friction, 
I^\ or tangential action of the plank 
against the stone, is always one third of 
the normal pressure, which is vertical and 
is evidently 360 lbs. (since the plank is a body at rest and therefore 
under a balanced system of forces, so that by moments about 
we find the normal pressure to be double the 180 lbs.). Hence 
the friction J^^ is = 120 lbs., and with regard to the plank is a 
force directed toward the right / but toioard the left^ as regards 
the revolving stone. The stone is a homogeneous right cylinder 
weighing 600 lbs., and of radius = 2', its geometrical axis being 
the axis of rotation. Radius of journals at 6^ is 1 in., and the 
mass of the projecting axle is neglected. The journal friction, 
F'\ tangent to the circumference, is to be taken as constant 
and as being i^oi the normal pressure, iV, on the journal. 




Fig. 102. 




WOEK AND ENERGY 11^ ROTARY MOTION. 89 

We now consider the stone free in Fig. 103. The forces 
acting are as shown, and the rotation clockwise 
with retarding angular velocity. 

Though the system is not a balanced one as 
regards rotation, since the moment-sum about C 
is not zero but = SMh^^ it is such as regards 
motion of the whole mass vertically ; for the 
centre of mass has a zero vertical acceleration, 
being at rest at all times. Hence we may put ^^^' ^^^' 

'2 (vert, comps.) = 0, and thus obtain, noting that the pressure 
N" is practically vertical, N" ^ G -^ N' \ and hence F" = 
960 -^ 20 = 48 lbs. 

To apply here the principle of Work and Energy, as expressed 
in eq. (2) of p. 144, M. of E., the range of motion considered 
being that occupied by the stone in coming to rest, we note, Fig. 
103, that G, JV\ and iY'' are neutral; that F' and F'' (both 
assumed constant) are resistances; and that there are no working 
forces. Let u denote the unknown number of turns made in 
coming to rest, ds' an element of the path of the point A, and 
ds'^ an element of the path of a point in circumference of the 
journal. With the J^oot, pound, and second, we have, therefore ; 

Work 

done 
onF' 



Work 



= / F'ds' = F' ds' = [120 X '^^ X 27r x 2] ft. lbs.; 

*y o *^ o 

Work ] rr^ rn 

done y= F''ds" = F'' ds''= [48 X ^X 2;r XyV] ft-lbs. 

on F" ) Jo Jo 

Since k^ for the cylinder (see p. 99, M. of E.) is = \r'' = 2 sq. ft., 
and the initial angular velocity is gd^ = 27r(180 ~ 60) = 18.85 
radians per second, we have 

The final kinetic energy is to be zero and the work of the work- 
ing forces is zero ; hence from eq. (2), p. 144, M. of E., 

= 4:S07ni + Sttu + [0 - (18.86)X18.6)]. . . (3) 

Solving this, we obtain u = 4.33 turns, for the stone to be 




90 NOTES AND EXAMPLES IN MECHANICS. 

brought to rest. Eq. (3) might be read : the initial K. E. is 
entirely absorbed in the work of over corning friction. 

(How would the time of coming to rest be determined ?) 
84. Work and Kinetic Energy. Motion of Rigid Body Paral- 
lel to a Plane. Numerical Example. — Conceive of a rigid body, 
or hoop, Fig. 104, consisting of two material points, JIf' and M^'\ 

so connected with a rigid, but 
imponderable, framework that 
the centre of gravity of the two 
material points lies at the centre 
of the circle formed by the outer 
edge of the frame. This outer 
circle, or wire, is to roll without 
Fig- 104. slipj)ing from a state of rest (cen- 

tre at o) to a position 6 ft. lower, vertically (centre at oi) ; q, in 
the figure, representing any intermediate position. The two 
masses are M' — 2 and if = 3 in the ft.-lb.-sec. system of units, 
their positions relatively to the frame being shown in the figure. 
Assuming that in passing position n the two masses are in 
line with the point of rolling contact m {M' uppermost), required 
their respective linear velocities, 'y/ and Vn^ at that instant. On 
account of the perfect rolling, each mass is at this fiual instant 
moving in a line ~i to the line connecting it with m, and Vy,' and 
Vn' are proportional to these distances ; 

i.e., vj : v^' :: 3 ft. : 1 ft. ; or ^^ = 'dv^'\ 
Throughout this range of motion {o to n) the acting forces 
are G, = 161 lbs., == the combined weight of the material points 
applied at the centre of the hoop (see AB in Fig. 104) and the 
normal and tangential components, JV and P, of the pressure of 
the fixed curved track or guide against the edge of the hoop at a. 
When any point of the hoop is in contact with the track, that 
point becomes the point of application of both JV and JP, and is 
at rest, being the one point of the hoop about which all others 
are turning for the instant (" centre of instantaneous rotation"), 
hence both iV and I^ are neutral forces. As the centre of hoop 
passes from q to q' through an element of its path, G does the 
work Gdh'y hence the total work done by G is Gfdh = 



ROTATING BODY AI^TD CORD. 91 

161 lbs. X 6 ft. = 966 ft.-lbs. of work. The initial K. E. of the 
whole body is = zero, and its final K. E., i.e., its (K. E.)^, is 

/. 966 ::= 4 V[18 + 3] - ; or <'^ == 92 ; 
whence v^' — 9.59 ft. per sec. ; and also v^ — 28.77 ft. per sec. 

[If a single material point were to slide from rest down a 
smooth fixed guide through this vertical height of 6 ft., its final 
velocity would be ( 1^2 X 32.2 X 6 =) 19.65 ft. per sec.]. 

If the hoop in the above example were to slip on any part of 
the guide, instead of rolling perfectly, the force P would no 
longer be neutral ; and if the edge of the hoop were to indent 
the surface without immediate and perfect recovery of form on 
the part of the latter, N and P would not be neutral. (See § 172, 
M. of E.) 

85. Numerical Example. "Work and Energy. Collection of 
Rigid Bodies. [Eead the ten lines under eq. (XYI), p. 149, M. 
of E.] — Fig. 105 shows a body mounted on a fixed horizontal 
axis containing its centre of gravity, of 

weight = 48.3 lbs. (hence its mass = ^^^'C^^^X Ixeollf!^ 
1.5, using ft., lb., and sec), its radius of 
gyration about the axis being Ic = 3 ft. 
This body includes a drum, as shown, 
from which a light inextensible cord 
may unwind until the point A of drum 
passes the position A' (the cord being "^^^^^^^"'fig. m 
firmly inserted), thus allowing the 10-lb. weight to sink vertically. 
C is a smooth fixed peg ; there is no friction at the journals. 
The figure shows the initial position (rest). The 10-lb. weight 
begins to sink, and the velocities to accelerate, while the cord 
simply unwinds until A reaches the unwinding point Z>, beyond 
which the point of insertion will follow the arc DA\ the portion 
of cord between it and C being straight. 

Kequired the (final) angular velocity oo^ of the rotating body 
when A reaches A' , 

Considering free this collection of rigid bodies (two masses 
and cord), knowing that all mutual actions between them, if 
normal to surfaces, are neutral, and assuming no internal friction, 




92 NOTES AND EXAMPLES IN MECHANICS. 

we Dote that the only forces external to the collection are the 
gravity- force of 48.3 lbs. (neutral because its point of application 
does not move) ; the normal reactions of the bearings against the 
sides of the journals (these are neutral because the path of the 
point of application is always -\ to the action-line of the force) ; 
the pressure of the peg against the side of the cord (neutral, for 
the same reason as that just mentioned) ; and the gravity-force 
of 10 lbs., which is the only working force. The path of the 
point of application of this last force is vertical and coincident 
with the action-line of the force, and the length of this path 
(which is equal to that of its own projection on the action-line) is 
equal to the total length of cord, s, that runs over the peg. Evi- 
dently s = 7tx^ft = 6.28 ft. (while A is passing to ^), 
increased by the difference between the distance 1^0 (considered 
straight ; a is verv small in this case) and the straight distance 

A'C; " 

i.e., s = 6.28 + [ VW + 2' - 14] = 8.406 feet. 
As both the initial and the final Kinetic Energy of the 10 -lb. 
mass are zero, and the initial K. E. of the rotating body is zero, 
the work 10 X 8.406 = 84.06 ft.-lbs., of the one working force, 
is entirely expended in generating the final K. E., ^go^MF, of 
the rotating body, 

.-. 84.06 = ioD^' X 1.5 X (3)^ and go^= 3.53 rad. per sec, 
which corresponds to gj^ -r- ^tt =^ 0.562 rev. per sec. 

As to the tension, S\ induced in the chord as the 10-lb. weight 
reaches its lowest point and begins to ascend, which is when the 
point of insertion of the chord in the drum passes through the 
point A (in its circular path), we note that at that instant, since 
the chord cannot stretch, the horizontal acceleration of the inser- 
tion-point, which is its normal acceleration at A' ^ must be the 
same*, as the vertical acceleration of the 10-lb. weight, and this is 
upward, since the downward velocity is slackening. Hence the 
resultant force, S' — G^ on (r at this instant is upward, and 
= (G -^ g) X {normal accel. of a jpoint moving with linear 
velocity = (y„ X 2 ft. in a curve loJiose rad. of cicrv. is 2 ft.), 
i.e., = {G---g)X {oon X 2 ft.)' -^ 2 ft. = 7.72 lbs. 
.-. S' = 10 + 7.72 = 17.72 lbs. 

* Approximate; the error is less the longer the distance -4' 6', compared with 
OA' , strictly true only when the peg is at an infinite distance toward the right 
(A somewhat similar approximation was made on p. 59, M. of E,, in consider- 
ing the connecting-rod of a steam-engine as infinitely long.) 



INTERNAL FRICTION — LOST WORK. 93 

At other parts of the motion the cord-tension is smaller. Let 
the student trace the remainder of the motion. 

86. Work of Internal Friction. — The student should note well 
that the work spent on the friction between two rubbing parts of 
the collection of rigid bodies under consideration is formed bj 
multiplying the value of the friction bj the distance rubbed 
through by one of the parts on the surface of the othei\ inde- 
pendently of the actual path in space of any point of either body ; 
for instance, in the examples of §§ 144 and 145, M. of E., the 
friction (F" = 400 lbs.) between the crank-pin and its bearing 
(in the end of the connecting-rod), the range of motion of the 
mechanism being that corresponding to a half-turn of the crank, 
is multiplied by the length of a half-circumference of the crank- 
pin itself, viz., 7tr'\ = 0.314 ft. (less than 4 inches), whereas the 
centre of the crank-pin describes a distance of ;rr = 3.14 ft. at 
the same time. 

87. Locomotive at Uniform Speed. — Kequired the necessary 
total mean effective pressure P^ in the cylinder of a locomotive 
on a level track to maintain constant the speed of a train whose 
resistance, li, at that speed is given. This resistance is due 
entirely to frictions of various kinds in the train and to the resist- 
ance of the air, and, the track being level, is equal to the tension 
in the draw-bar immediately behind the locomotive. Fig. 106 
shows the collection of rigid ^ .^^ 

parts forming the locomotive /^ \^ -^^ ^ |Q <-2a->^ 

alone and the forces external ,«— tf-^) ^^-^.^j 5 ^^ ' ;]| ^p^ -►? 
to them, all mutual frictions V ^^^^f^^/s^£^ H I W I l i 
being disregarded. For sim- ^^dI/S^ vTV \Yty 



plicity consider that there is C3 n;5 uy l^ - ^-^Z '> 
only one cylinder and piston ; fig. loe. 

and take, as the range of motion of the parts, that corresponding 
to one backward stroke of the piston, i.e., to a half-turn of the 
driving-wheels. The external forces are : /^^ on the piston ; an 
equal P^ on the front cylinder-head ; Y, Y, the pressures of the 
rails against the truck- wheels ; S^ S, those of the track against 
the driving-wheels, which are supposed not to slip on the track; 
Gf the weight of the locomotive ; and jR, the tension in the draw- 



94 NOTES AND EXAMPLES IN MECHANICS. 

bar. Of these, the Z's and the /5"s are neatral for reasons given 
in the example of § 84 (perfect rolling); G also is neutral, the 
elements of its path being always ~\ to the action-line of the force ; 
j^ is a resistance, overcome through a distance = nr ; the P^ on 
the front cylinder-head is a working force, its point of application 
describing a horizontal path of length = nr ; while the P^ on 
the piston is a resistance whose application-point moves forward 
in absolute space a distance =l 7tr — 2a. The kinetic energy of 
the mechanism being the same at the end as at the beginning of 
the stroke, the K. E. terms cancel out from the equation of work 
and K. E., leaving 

P,7tr - PlTtr - 2a) -Ji7tr = 0; .: P„ = ^, 

or P^ = the draw-bar resistance multiplied by the half-cir- 
cumference of the driver and divided by the length of one 
stroke. For example, if a 200-ton train offers a resistance 
of 10 lbs. per ton (in the draw-bar) at a speed of 20 miles 
per hour, then wdth 2.5 ft. radius for the drivers and 1 ft. 
radius for the crank-pin circle, we have P^ = 7857 lbs. ; i.e., with 
two cylinders, 3928 lbs. for each piston. Suppose each piston to 
have an area of 100 sq. in., then the average amount by which 
the steam-pressure on one side should exceed the atmospheric 
pressure on the other is 39.28 lbs. per sq. in. As steam is ordi- 
narily used expansively when the train is under full headway, this 
means an initial steam-pressure at the beginning of stroke of per- 
haps 80 or 90 lbs. per sq. in. above the atmosphere (as a roughly 
approximate illustration). Above, we have assumed the drivers 
not to slip on the rails. Slipping will not occur ordinarily if 
P is less than about one fifth of the total weight resting on the 
drivers. 

88. Locomotive Tractive Effort at Starting. Track Level. — 
When the train is once in motion the steam-pressure on the 
piston in the latter part of the stroke is much smaller than 
the average, so that the speed slackens temporarily, the great 
mass of the train acting as a fly-wheel. To start, however, this 
steam-pressure must reach a certain minimum amount which we 



MECIIAiq-ICS OF THE LOCOMOTIVE. 



95 




Fig, 107 



Fig. 109. 



call P, For instance, with one engine at its dead-centre, tlie 
duty of starting devolves on the other alone, which is then in the 
mid-stroke position (nearly), its crank being vertically over (or 
under) the driver-axle, and the horizontal component of the pull 
on the crank-pin is = P. Fig. 107 
shows the driving-wheel as a free 
body, the vertical components of 
the acting forces being omitted, as 
having no moments about 7>, the 
point of contact with the track. 
(Steam is now pressing on the left 
face of the piston, not shown, and 
on the hinder (left end) end of cylinder-head (see Fig. 106 for 
direction of motion, etc.). The action of the driver is that of a 
lever whose fulcrum is at D (no slip). P' is the horizontal press- 
ure of the locomotive-frame against the driver-axle, as induced by 
the pull P ; and by moments about D we have P'r — Pi^r -\- a) ; 

i.e , P' = P -\ — P. Considering now the locomotive-frame in 

Fig. 108, we lind a forward force 
P' at A, the bearing of the 
driver-axle ; the draw-bar resist- 
ance R, backward ; and a back- 
ward steam-pressure =^ P on 
Fig. 108. the hinder cylinder-head at B ; 

whence, for equilibrium (i.e., more strictly, for a very small 

fr\ 
acceleration), P' — P — R, and finally P — \~)R^ as the neces- 
sary total effective steam- pressure on the one piston to start the 
train w^hen the draw-bar resistance is R (neglecting the resist- 
ances in the locomotive itself). (N^o vertical forces sliown.) 

Similarly, if the crank-pin were vertically under the axle, 
as in Fig. 109, we obtain for the pressure at axle (induced by 

'a^ 




P) the value P' ^ P — [t;)^^ [since, from moments about P, 

P'r = P{r — <3^)]. Hence, as regards the locomotive-frame, we 
now find a hachward force P' at the axle-bearing ; but there is 



96 NOTES AND EXAMPLES IN MECHANICS. 

now a forward steam-pressure, P, against the front cjlinder- 
Lead, so that P — P' = B^ which after substitution gives us 

P = l-jP, the same as before. 

If the wlieels are not to slip, the horizontal action of the rail 
on the driver, viz., P'\ Figs. 107 and 109, must not exceed the 
ultimate friction, P, or about one fifth the weight on the drivers. 

But P'' = -P, so that P" =: P ; hence P should not exceed P. 

89. Appold Automatic Friction-brake, or Dynamometer of 
Absorption.— Fig. 110 shows the Appold friction-brake, which is 

automatic by reason of its 
•' compensating lever," PC. 
The brake is formed of a 
ring of wooden blocks, con- 
nected by an iron hoop, 
whose two ends are pivoted 
to the lever, as shown, at 

Fig. 110. ' Fig. 111. C. The pullcj PFrcVOlvCS 

within. If the friction is insuflScient to keep the weight G raised, 
it sinks and the end B of the lever presses against the fixed stop 
PK^ thus tightening the hoop, increasing the friction, and raising 
G (aud vice versa when the friction is too great). 

Fig. Ill shows the Appold brake as a free body. The normal 
pressures of the pulley against the wooden blocks have no mo- 
ments about (7, while the tangential actions (i.e., the frictions) 
have a common lever-arm, == a, about C. Hence, by equilibrium 
of moments, Pa =z Gh — Pc, where Pis the sum of the frictions. 
Evidently the pressure P at the end of the lever should be 
known, for accuracy. 

(See correspondence in the London Pngineer^ from IS^ovem- 
ber, 1887, to March, 1888.) 

Hence if v is the velocity (ft. per sec. for instance) of a point 

in the pulley-ritn, the power absorbed is Z = Fv = [ \v. 

90. The " Carpentier" Dynamometer of Absorption. (Exhibited 
in Paris in 1880.) — This is automatic, like the Appold brake, but 




THE '' CARPENTIER " DYNAMOMETER. 



97 




Carpentier 
k Dynamometer 



instead of automatically altering the tension in the strap to keep 
the weight " floating," it changes the arc of 
contact A B until the friction is so altered 
(= i^') as to equilibrate the weight G' with 
help of the smaller weight G. See Fig. 112. 
iV^is a pulley keyed upon the shaft of the 
motor to be tested, and therefore moving 
with it. A weight G is fastened to a strap 
BA C against which the pulley N rubs, but 
the upper end of this strap is attached to a 
block C^ which is a rigid part of another || 
pulley, or disk, M, heyond the pulley W. ^P. 
The disk M is loose on the shaft, the block 
C projecting over the face of the pulley iT. The strap DG' is 
attached to the pulley M 2X D and carries a weight G' which we 
at first assume to be supported by a fixed platform 'k^ . When 
the shaft begins to turn (see arrow), and the tension in CA 
occasioned by the friction due to the arc AB and weight G is 
greater* than G\ then G' will begin to rise, the disk M turning. 
But, as M turns, the uppermost point of contact of the strap 
AB on the pulley iV^ moves to the right ; i.e., the arc of contact 
AB becomes smaller, with a consequent reduction of the friction 
between N and the strap, so that after a little the tension in CA^ 
pulling on C^ is just sufficient (= S') to keep the weight G' at rest 
at some point k' . When this state of equilibrium is reached, we 
have, by moments, for the equilibrium of the strap, S'r^^F[r-\-Gr'^ 



and from that of the disk, S'r = G'r' ; i.e., F' = - 

that U V = velocity of the pulley-rim, the power 

GV 1 

— G V, (ft.-lbs. per second, for instance.) 



GV - Gr 



r 



so 



F'v. 



=c 



91. Numerical Example in Boat-rowing. — As an illustration of 
the relations of the quantities concerned in a simple, but typical, 
case of propulsion on the water, let us suppose, in the problem of 
Fig. 166 of p. 161, M. of E., that the distances from theoar-bandle, 
A^ and oar-lock, C^ to the middle of the blade are 9 ft. and 6 ft., 
respectively ; and that a pull of 20 lbs. is exerted on A. The 
pressure at the oar-lock is then 30 lbs. ; and that of tne blade 

^ Or, rather, its moment greater than that of G' . 



98 NOTES AND EXAMPLES IN MECHANICS. 

against the water, 10 lbs. Hence the boat is subjected to two 
forward oar-lock pressures of 30 lbs. each ; to two backward 
pressures against the foot-rest, of 20 lbs. each ; and to some back- 
ward resistance R of the water against the hull (i? depending on 
the square of the velocity ; R = zero if there is no velocity). 

The difference between the oar-lock and foot-rest pressures is 
a forward force of 20 lbs. ; and if the velocity of the boat at the 
beginning of the stroke is such that ^ is = 20 lbs., the effect is 
to barely maintain that particular speed while the 20 lbs. pressure 
is acting on each oar-handle, (and R remains constant also.) If 
R is smaller than 20 lbs. the velocity will be accelerated, and R 
will increase ; if larger, the velocity diminishes (and R also), and 
of course will diminish at a much more rapid rate when the oar 
is lifted from the water. 

For an ordinary small skiff (with pointed stern as well as bow) 
containing one person, the water-resistance R is (roughly) about 
one-half pound for each sq. foot of the wetted surface of the hull, 
when the velocity is 10 ft. per sec. / for other velocities, as the 
square of the velocity. 

If in above case each oar-handle, while under the 20 lbs. press- 
ure, passes through a distance AJE = 3 ft., measured on the loat, 
and the blade slips backward in the water an absolute distance 
5j of 6 inches (say), = J ft., the absolute distance passed 
through by the boat will be (from the geometry of the figure) 
5.5 ft. The work spent on slip is 2 X 10 X J = 10 ft.-lbs. ; so 
that, of the work, 2P . A^ ^ 2 X 20 X 3 = 120 ft.-lbs., exerted 
by the oar-handle pressures, relatively to the hoat (see fourth line 
of p. 161, M. of E.), 110 ft.-lbs. remain for the work of over- 
coming the resistance, ^, and increasing the K. E. of the mass 
of boat. R is overcome through the distance ^3 == CD^ = 5.5 ft., 
so that if all of the 110 ft.-lbs. are spent on R (i.e., if the velocity 
is to be maintained constant), we must have 110 =z R x 5.5, i.e., 
^ = 20 lbs., as proved above. [In order that R may have this 
value with a small skiff the velocity must be (roughly) about 11 
or 12 ft. per sec, at the beginning of stroke.] 

'Note that the absolute distance through which the two 20-lb. 
oar-handle pressures work is 5.5 -|- 3 = 8.5 ft. But, of the abso- 
lute work done by them, viz., 2 X 20 X 8.5 = 340 ft.-lbs., in the 



NUMERICAL EXAMPLE IN BOAT-ROWING. 99 

Stroke, the amount 2 X 20 X 5.5, = 220 ft.-lbs., is absorbed in 
overcomiDg the foot-rest pressures through 5.5 ft., the remainder 
being an amount 2 X 20 X 3, = 120 ft.-lbs., = 2F . AF, to be 
spent on the work of slip, of liquid-resistance, and change of 
K. E. (if any). 

92. Remarks on the Examples of § 155, M. of E. — In the first 
example the work to be computed is that done by the tension 
P, in the draw-bar, considered as a working force acting on 
the train behind the locomotive. If the velocity were uniform, 
a value 10 X 200 = 2000 lbs. would be sufficient for P ; but as 
the velocity is to be increased, the " inertia" of the train is brought 
into play and the amount required is three times as great in this 
instance. In Exaviple 2 P has the same significance. 

As to Example 3, multiplying the 15,000 lbs. resistance by 
the speed reduced to ft. per sec. will give the power in ft.-lbs. 
per sec. Dividing this number by 550, we obtain 461 H. P. (see 
p. 136, M. of E.). 

In Example 4 the resistance is greater than before, in the 
ratio of (10)' to (15)', i.e., it is 21- times 15,000 lbs. The distance 
through which it is overcome per second being 1|- times its former 
value, we find the power spent on water-resistance to be (in ft.-lbs. 

per sec.) 2^ X li X 15000 X -q^^j-q^ ', which divided by 550 

gives 1556 H. P. (That is, the respective powers are as the 
etches of the speeds.) 

Example 5. Since 80 per cent of the power, Z, of the work- 
ing force (steam-pressure) is to be 461 H. P., we write 0.80 X L 
= 461, and obtain Z = 576 H. P. 

Example 6. If the thrust or pull of the connecting-rod of the 
engine on the crank-pin be resolved at every point into a tano-en- 
tial and a normal component, Tand N (Fig. 
113), we note that Z is a working force and 
iV neutral. Hence at this point in the line 
of transmission of power we can ascribe all 
the power to the force T. T is variable, and 
by its average value, Z^, we mean a value whose product by the 
length of the circumference described by the crank-pin shall be 
the same amount of work as that actually done by the variable 




100 



NOTES AT^D EXAMPLES IN MECHANICS. 



T per revolution. ISTow 461 H. P. means 253,000 ft.-lbs. of work 
per second, which divided by 1.0, the number of turns per sec, 
gives the work done bj T in each turn. Hence Tyn^n X 1.5 
should = 253,000; or T,^ = 26,890 lbs. 

Mcample 7. Fig. 114. At we have the sphere in its initial 
position, the forces acting on it being its 
own weight G (a resistance), and the two 
components, iVand T, of the pressure of 
the inclined plane against it. Since there 
is no slipping (i.e., perfect rolling), both JV 
and T are neutral (see § 84). 

Let v^ = initial velocity of the centre 
of gravity, and od^ the initial angular ve- 
locity, s' sin 30° is the unknown height through which the centre 
will rise before the velocity becomes zero. 

G v"" 
The initial K. E. of translation is — • -7^ : 

g 2 

ioD,'Mk,\ Now V, = GD,7' and k; = |/ (p.- 102). 
(XY), p. 147, 




Fig. 114. 



and of rotation, 
Hence hj eq. 



-Gs\^ 



o-6^r 



3x3-1 

2 X 32.2 



(1+1); or / = 0.39 ft. 



93. Efficiency of a Wedge. Block on Inclined Plane. — In the 
numerical example 3 of p. 1Y2, M. of E., it is to be noted that 
the " efficiency" of the mechanism, or ratio of the work usefully 
employed in overcoming Q, to that exerted by the one working 
force P, is 57.6 -^ 153.6 = 37.5 per cent. 

Fig. 115 shows as a free body the block mentioned in prob- 
lem 6, p. 172, M. of E. We are to find the 
force P in the given action-line, such that a 
uniform motion (equilibrium) can be maintained 
ujp the plane. The action of the plane on the 
block is represented by the normal pressure N 
and the tangential action, or friction, /W. [The 
student should not assume thoughtlessly that N must = G cos jS 
(the normal component of G\ for the unknown P, not being 





EOLLING RESISTANCE. 101 

parallel to the plane, has an influence in determining iT; here it 
tends to relieve the pressure on the plane.] The forces being 
balanced, ^X, and ^ Z, = ; 

whence P cos a —fJV— G sin /3 — 0; 

and P sin a: + i\^ - 6^ cos y5 = ; 

and finally F = — ^ , ' ;. . — — (1) 

•^ cos a -f-j sin a ^ ' 

Problem 7 calls for the value of P if the motion is to be 
doivii the plane ; other things as before. Fig. 
116 shows the change. The friction acts in a 
contrary direction ; jP will be smaller ; JV, larger. 
Solving as before, we finally have 
6^ [sin yg -/cosyg] 
cos a —f sin a ' • ' \"J ^^^^ jjg 

We note here [eq. (2)] that ior fi = ct) = the angle of friction, 
P r= ; and that if/* is large and /? small (< 0), P may be nega- 
tive, i.e., its direction may need to be reversed, to aid the body 
down the plane. 

94. Work Absorbed in Rolling Resistance. — With perfect roll- 
ing of a smooth rigid wheel upon a straight, fixed, smooth, and 
incompressible rail, the pressure jR of rail on wheel is a neutral 
force as regards work (call this ^'perfect rolling''') ; for its point 
of application, at the foot of the perpendicular let fall from the 
centre of wheel on the rail, is motionless so long as it is the point 
of application. But with an inelastic, compressible rail, the 
point of application is a little ahead (distance = h) of the foot of 
that perpendicular, and it results, therefore, that for every ds 
moved through by the centre of the wheel the rail-pressure is 

overcome through the small distance \--\ds\ so that when the 

wheel-centre (or any point of the car-frame) is passing through 
the distance s (parallel to rail), the work done on the rail-pressure, 

P, is not zero, but = R . -s. It is convenient, therefore, in 



102 NOTES AND EXAMPLES IN MECHANICS. 

applying the principle of work and energy to the case of a car, 
to consider that the rolling is j^erfect and that the work actually 
due to rolling resistance is occasioned by the overcoming of an 

imaginary backward force, —7?, applied directly to the car-frame, 
or centre of axle of wheel ; since the product -B s is eqnal to 

that of R by [—s]- This imaginary force is what is referred to 

in the foot-note of p. 189, M. of E. 

95. Examples of § 173, M. of E. — In Example 2, since slipping 
is impending, the backward tangential action, or force, of the rail 
on each wheel must be T, = (0.20 x 5000) lbs. ; and since its 
moment, Ti\ about the wheel-axis must balance that, Fr, of the 
friction i^ between brake and wheel-rim (neglecting the moment 
of the small axle-friction), this friction must be i^= 1000 lbs. 

When the car-frame moves through the unknown distance s 
before coming to rest, the friction i^(an internal friction) is over- 
come through a relative distance measured on the wheel-tread of 
length = s (because the wheels do not slip and the brake-shoe 
presses on the same rim as that which rolls on the rail ; whereas, 
if the shoe rubbed on the rim of a drum on the same axle, the 
circumference of the drum being (for example) one half that of 
the wheel-rim, the relative distance rubbed through would only 
be one half of s. 

Gravity being neutral and both axle and rolling resistances 
being neglected, we have, from eq. (XVI), p. 149, M. of E., (with 
foot, pound, and second,) 

0-8 X 1000X5 = 0-^^^(80)^; .-. 5 = 496 ft. 

In Example 3, although the track is on an up-grade, or 
inclination «', with the horizontal, the angle is so small that 
60 -^ 5280 may be taken as either its nat. tang, or its sine, at 
will. The weight 40,000 lbs. is here a resistance, being raised 



EXAMPLES IN DYNAMICS. 103 

through a height of 1000 ft. X sin a, = 11.35 ft, and both roll- 
ing and axle resistances are neglected. Hence 

- 8i^ X 1000 ft. - 40000 lbs. X 11.35 ft. = - ^ • -322"^^^^'' 

or i^==439 lbs. 

Example 4 differs from the preceding only in the introduc- 
tion of two more items of negative work. The work of rolling 
resistance is ascribed to the overcoming of an imaginary force 

' 02 \ 

*Tj-^ X 5OOOI lbs., = Q.Q^ lbs. for each wheel (applied to car- 
frame), through a distance ot 1000 feet. 

The axle-friction at the journal of each wheel is (0.036 X 5000) 

lbs. — 180 lbs. (taking the coefficient near the top of p. 190, 

M. of E.). A point in the circumference of the journal rubs 

'1 5\ 
through a distance of ( -^j X 1000 ft. = 100 ft. Hence, finally, 

- Si^X 1000 - 40000 X 11.35-8x6.66 X 1000-8 XlSO X 100 

= ^-|-W(^^)" ^•^•' ^"^=^l^^bs. 

96. Examples in Dynamics. (Statements on p. 194, M. of E.) 
— Exam2?le 1. Fig. 117 shows the end of the shaft in question. 
The friction on the side, or "axle friction," is 
due to the pressure of 40 tons, = N' ; i.e., 
F ^fN' =■ .05 X 80,000 = 4000 lbs., and is 
overcome through a circumference of tt x 1 ft., 
= 3.14 ft., each revolution. The w^ork done per 
revolution on the friction at the base (due to the 10 tons) is the 
same as if all concentrated at a distance = f of the radius 
from the centre, and is therefore (.05 X 20000) X (Itt X 1) — 
(1000 lbs.) X (2.094 ft.). Hence the lost work per second, i.e., 
iXi^jpoxoer^ absorbed by friction, in ft. -lbs. per sec. ^ is 

f^[4000 X 3.14 + 1000 X 2.094] = 12211 == 22.2 H. P, 




104 



NOTES AND EXAMPLES IN MECHANICS. 



Example 2. Fig. 118. (This form of frictiun-brake is some- 
times used in testing motors.) Since slipping ac- 
tually occurs^ we use eq. (3) of p. 183, M. of E., 




^.= ^>/«; i.e,> = ]og, r|]. 
S^ = 20 lbs., S, = 10 lbs., with a ^ 



Here w^e put 



Tt, and obtain 



/= [2.302 X log,o(2.0)] - TT, zzz 0.22. 
Fig. 118. As to the power, Z, expended on the friction, we 

note that each element of the friction is overcome through a dis- 
tance V every second, where v is the linear velocity of the pulley- 
rim; i.e., Z = Ev, denoting the sum of the elementary frictions 
by E These elementary frictions are not parallel, but we note 
that each (being tangential) has the same lever-arm, = 9 inches, 
:= r, about the centre of the circular curve, so that when the 
curved part of the strap is considered free, and moments are 
taken about that centre, we obtain ZV = S^r — S^r (the normal 
pressures on the cord have zero moments) ; hence E — Sn — ^o 
and the power in question =z Z = [S^ — So'jv. Therefore 

Z = [(20 - 10) lbs.] X [(27r X^X ^V") f*- per sec] = 78.5 

ft.-lbs. per sec. ; = 0.142 H. P. 

Should the pulley be made to turn the other way with the 
same speed, the weight of 20 lbs. (now = S^) remaining the same, 
more power will be required. Assuming the coefficient of fric- 
tion between strap and pulley to remain unchanged, the spring- 
balance will now read 40 lbs. tension (= S^), since ef*^ = 2, so 
that Sn — S„ = 20 lbs. instead of 10 as before, v being the same 
(7.85 ft. per sec), the power will be double that previously 
found ; viz., Z now = 0.282 H. P. 

Examjjle 3. Fig. 119. It is assumed that the pressure on the 
journals, or axle, is due solely to the weight, 6', 
of the stone ; i.e., that the caps of the bearings are 
not clamped in contact with the journals; other- 
wise the friction would be '> f G^ for axle-friction. 

The stone being considered free, we note that 
its weight and the iionnal component of the reac- fig. i 

tion of the bearing are neutral forces, the friction, or tangential 




EXAMPLES IN DYNAMICS. 



105 



component, /''6'^, is a resistance; and that there are no working 
forces. 

The initial K. E.*^ of the stone is ^oo^^ — Ic" {oo^ being the initial 

angnlar velocity) ; its final K. E., zero. Hence the initial K. E. 
is all absorbed in the work of friction. Since the number of 
turns in coming to rest is 160, the total distance through which 
the friction at the circumference of the journal is overcome is 
160 X ;r[1.5 ~ 12] = 62.85 ft. With t^Qfoot and second, go, = 
27r X -W" — ^^ radians, and g = 32.2. Therefore, from the 
Principle of Work and Kinetic Energy [eq. (XY), p. 147, M. 
of E.], 

(1 /1Q\2 

i(^;ry 3^2(12 j ^f'^ X ^^-^^ 5 and .'. f = 0.039. 
Example 4. To move A horizontally with an acceleration 
= 15 (foot and sec.) would require a horizontal force = mass X 

2 
-— X 15 == .93 lbs. But the horiz. compon. (friction) 



ace. = 



of the action of B on A is only — fN = 0.3 X 2 = 0.6 lbs., 
which is < .93. Hence A will not keep abreast of B^ but will 
gradually fall behind B. 

97. Brake-strap, Lever, and Descending Weight. Numerical 
Example (Fig. 120).— The weight Q of 600 lbs. is to be let down 
without acceleration, the rope un- 
winding from a drum of 1 ft. 
radius. On the shaft of the drum 
is keyed a pulley of 2 ft. radius, 
the friction on whose rim, due to 
its rubbing under the encircling 
stationary strap, can be varied in 
amount by a force P exerted on 
the lever AB. It is required to 
compute a proper value for the 
pressure P in the present instance 
i<d prevent acceleration of the 600-lb. weight, the coefficient of 
friction of the strap on the pulley being assumed to be/'= 0.30. 

* Kinetic Energy. 




Fig. 120. 



106 NOTES AND EXAMPLES IN MECHANICS. 

The strap covers three quarters of the pulley-rim (i.e., n = |;r). 
See figure for other dimensions. 

If Q sinks without acceleration, the tension in the vertical 
part of the rope must be 600 lbs., and the rotation of pulley be 
uniform ; hence moments must balance for the pulley, drum, and 
shaft ; so that (with the foot, pound, and second) 600 x 1' = 
j?^X 2', where i^'is the sum of the requisite frictions (tangential 
forces) of strap on pulley ; i.e., F = 300 lbs. 

But from the equilibrium of the curved portion of strap, by 
moments about centre of the curve, S^x 2' — SnX 2'-\-Fx 2'= 0. 
Sn is the tension in the vertical straight part of the strap ; S^ , 
that in the horizontal. S^ is greater than S^ and bears to it the 

relation Sn = S^e^" ; or, ^/tt = loge (-rr). That is ^^ee p. 184, 

M. of E.), {Sn : So) = the number whose common logarithm is 
(.45);r X 0.43429 = 4.12. Combining this with F= S^- S, 
(see above), we have (4.12 — l).^^ == 7^ = 300 lbs.; whence 
S, = 300 ~ 3.12 =: 96.15 lbs. ; and Sn, = 4.126^, = 396.13 lbs. 
The requisite force P is then found by noting that for the 
equilibrium of the lever, the moments of the three forces jS^, S^j 
and i^, must balance about the fulcrum A ; i.e., 

Px4^=:^n xi + ^oX4; or P = i{Sn + S,) = 61.5lhs^ ' 
a pressure easily applied by one man. 



CHAPTER YII. 

Mechanics of Materials and Graphical Statics. 

98. Intensity of the Shearing, and of the Normal, Stress on an 
Oblique Section of a Prism under Tension. (This treatment may 
be more readily understood than that now given in § 182 of M. 
of E.) — From the prism under tension in Fig. 193, M. of E., con- 
sider bv itself a portion shown in 
Fig. 121, between the right section >v 
ICJS^ and any oblique section, ML. 
The area of the right section being 
J^ and the intensity stress per unit 
area of that section being p, Fp /,^ ^ 

expresses the total stress on the ^^^- ^^^• 

right section. The total stress on ML is also of course = F'p, 
Its component P" normal to the plane ML is evidently P" = 
Fp sin a. This is the total normal stress on this oblique plane 
ML. But the area ML over which this normal stress is dis- 
tributed is not = F^ that of the right section of the prism, but 
= {F -^ sin oi). Hence to obtain the normal stress on ML per 
unit of area^ i.e., the intensity of normal stress on J/Z, we must 
divide P" by {F ~- sin a) and thus obtain : 

Normal stress, per tinit of] . ^ ,^.f 

area,, on ohlique section ) -^ "" ' * • w 

Similarly, the other component, P^ (of the total stress on 
ML), which is tangential to ML, is in amount = Fp cos cr, which 
is the total shearing stress on ML. To obtain the intensity of 

107 



108 KOTES AND EXAMPLES IN MECHANICS. 

this shearing stress, we divide P' by the area {F -^ sin a) of ML, 
over which P' is distributed, and obtain : 

Shearinq stress, per unit \ . , ^ , 

^ . , . , . r = » . sin « cos «'. . (2) 

oj area^ on ooiique section ) ^ ^ ^ 

In these two equations jj is an abbreviation for P -^ F^ and 
a is the angle that the oblique plane ML makes with the axis of 
the prism. 

The reason for ascertaining the stress jper unit area in any 
case is, of course, that the safety of the material depends upon it 
and not simply on the total stress. 

99. Spacing of Rivets in a Built Beam. — The statement in the 
middle of p. 293 of M. of E. that '' The riveting connecting the 
angles with the flanges (or the web with the angles), in any 
locality of a built beam, must safely sustain a shear equal to J 
(the total shear of the section) on a horizontal length equal to the 
height of the web^'' may be most directly utilized as follows : 
Imagine the horizontal continuity of the web to be broken and 

consequently a vertical seam ren- 
dered necessary, as shown in Fig. 
122. 

Then, whatever spacing of rivets 
would be necessary in this ideal seam 
can be adopted in the real horizontal 
^^^' ^^^' seam made by riveting together the 

web and angles, the rivets being considered to be in double shear 
(or single) in the ideal case, if so in the actual. For examjjle^ 
taking the data of the example on p. 294, M. of E., there must 
be enough rivets in the vertical seam, of length = /?„ = the 
height of web, to safely stand the total shear of J = 40,000 lbs. 
Since each rivet can safely endure a shear of 9000 lbs. in double 
shear (see p. 294), the number of rivets required in the height of 
web would be 40000 ~ 9000 = 4.44 ; i.e., they should be spaced 
4.5 in. apart since the height of web is 20 in., and 20'^ ~ 4.44 = 
4.5^'. 

But since the pressure on the side of each hole is limited to 
i>470 lbs. (see p. 294), on this basis the number of rivets in height 



.'-": 


^^-^^"--— 


-^-- 


(L_C 


^ ^^-IR 


'^r^' 


Tj 


n Y 




--^ ^ 


\ \y^ 



MECHANICS OF MATEKIALS. 



109 




Fig. 123. 



of web should be 40,000 -^ 5470 =: 7.2, which implies putting 
them at a distance apart, centre to centre, of 20 -f- 7.2 = 2.7 in. 
This spacing, therefore, should be adopted in the horizontal seam 
between web and angles at this part of the beam (near abutment). 
100. I-beams treated without the Use of the Moment of Inertia. 
— The assumption is so frequently 
made (for simplicity in treating the 
web) that the web carries all the 
shear, that the corresponding as- 
sumption that the two flanges carry 
all the tension and compression is 
also sometimes used to simplify the 
analysis. With thin webs the re- 
sults obtained are accurate enough 
for practical purposes. For ex- 
ample, suppose a horizontal I-beam to rest upon supports at its 
extremities and to bear several detached loads. To find the 
tensile or compressive stress in the flanges at any right section as 
m, consider free the part of beam on the left of that section (see 
Fig. 123). i? is the abutment reaction ; P^ and P^ the loads 
between m and A, Considering the total compression P' to be 
uniformly distributed over the area of the upper flange, and 
hence (for geometrical purposes) to be applied in the centre of 
gravity of that flange, and similarly the total tension P" \x\ \X\q 
centre of gravity of the lower flange ill' = vertical distance be- 
tween), while the web carries the shear alone; we obtain (by 
taking moments about the intersection of «/and P")^ 

p^a^ _p p^a, + P'N = Ra ; 

from which P' is found. Evidently P"z=^ P\ since they are the 
only forces having horizontal components. Let i^' = the area of 
either flange, then the stress per unit area in either flange is 

P' . 

^' = -^. If this result is too large for safety the flange-area 

must be increased. 

By this method, the proper amount of sectional area can be 
computed for the flanges at each of several sections of a beam to 



110 NOTES AND EXAMPLES IN MECHANICS. 

carry a specified loading, and a beam of " uniform strength" be 
designed. In such a case, however, the weight of the beam itself 
cannot be estimated in advance, but after a gradual tentative 
adjustment can be taken into account in a satisfactory manner. 

(The upper flange being in compression may need to be 
braced or latticed with those of the accompanying parallel beams 
to prevent horizontal buckling.) Plate girders of variable section 
are designed on this general principle, the flange-area being 
increased toward the middle of the span by riveting on additional 
plates. 

101. Load of " Incipient Flexure" of Long Slender Columns. — 
The result brought out in eq. (8) of p. 366, M. of E., that in the 
case of a long, slender^ round-ended, prismatic column, no flexure 
at all takes place until a definite value for the load is reached, and 
that, with that value, any small deflection whatever can be main- 
tained, may be arrived at quite rationally as follows, without 
intricate analysis : 

Let the horizontal bed-plates of a testing machine. Fig. 124, 
be advanced toward each other until the slender round- 

k\ ended rod AB between them is deflected a small amount 
/ 1 z=z a at the middle. If p is the radius of curvature of the 
\\\ elastic curve at i?, we have (from free body R , . A^j 



57 ^ 




^ EI 


— = Pa', 


or 


P=—, 


P 




ap 



(This radius of curvature, p, is nearly — to that of the 
circle determined by the three points B^ i?, and A {^ 
of it). 

Now let the plates be separated until the deflection is 
only = a' ; call the new radius of curvature, p' . Then the 

EI 

Fig. 124. ucw valuc of the force or pressure at each end is P' = —j—,. 

But, for this very flat curve, as the deflection is diminished, the 
radius of curvature changes in an inverse ratio ; * i.e., p' \ p',\a\a' \ 
or, ap = a' p' , Therefore P = P' \ i.e., the jpressnre induced ly 
the elasticity of the rod against the plates does not diminish loith 
a diminishing defljectioyi, hut remains constant. Or, conversely, 
if a less force than this is applied to the column, no deflection 

* Strictly, the relation is 

a(2p — a) = [^AB]', = practically a constant. 



TESTS OF WOODEN POSTS. Ill 

takes place ; while if an actual load, whose weight is greater than 
the above force, be placed on the column, the upward pressure 
against it due to the elasticity of the column (as the latter bends) 
being always less than the weight (unless perhaps when the de- 
flection becomes extreme), the load sinks with an accelerated 
motion and the column finally breaks (since with increased deflec- 
tion the stress in the outer fibre is augmented). 

In most cases in practice, columns are not sufiiciently slender 
to bear out all the above-described phenomena, but enough has 
been said to show that while with a horizontal beam the defiec- 
tion is nearly proportional to the load applied (within elastic 
limit), such is far from being the case with a column. 

If a piece of card-board cut from a visiting card, and about 
half an inch wide by three or four inches long, be pressed end- 
wise with the finger on the scale-pan of a letter-scale, the value of 
the pressure corresponding to diflerent deflections can be easily 
noted and the above claims roughly verified. 

102. Recent Tests of Large Wooden Posts — Prof. Lanza, of 
tiie Massachusetts Institute of Technology, Boston, has made 
tests on yellow-pine flat-ended posts, giving results as follows : 
Highest breaking stress 5400 lbs. per sq. in. ; lowest, 3600 ; 
average, 454^1:. 

These yellow-pine posts were nearly cylindrical in form and 
almost all of them 12 ft. in length (a few 2 ft. long); diameters, 
from 6 to 10 in. 

With white oak posts, flat-ended, and of about the same sizes 
as the former, the highest breaking stress was 4600, the lowest, 
3000 lbs. per sq. in. ; with exception of two which reached 6000. 
In all these cases of pine and oak posts failure occurred by direct 
crushing, lateral deflections being inconsiderable, showing that 
all were practically '' short hlocksP 

Eight separate tests were made with the load applied eccen- 
trically, about two inches off the centre, the result being to 
diminish the strength by about one third. All these posts had 
been in use for years and were well seasoned. Each had a hole 
about 2 in. in diam. along the axis, from end to end. 

Othei' tests have been made at Watertown, Mass., with the 



112 NOTES AND EXAMPLES IN MECHANICS. 

Government testing macliine on timber columns, of rectangular 
sections, mostly about 5 by 5 in., and 7.5 bj 9.7 in. ; with a num- 
ber 5 by 15 in., and 8 bj 16 in. Their lengths ranged progress- 
ively from 15 in. to 27 ft. Flat-ended supports. From these 
tests Mr. Ely concludes that if the breaking load in pounds be 
expressed as P = FC^ where F is the sectional area in square 
inches, one of the following values for C should be taken accord- 
ing to the kind of timber and the ratio of the length I to the 
least side, 5, of the rectangular section ; thus : 

For white pine : 

For (^ -^ ^ = to 10 ; 10 to 35 ; 35 to 45 ; 45 to 60, 
Make C= 2500; 2000; 1500; 1000. 

For yelloio pine : 

For Z-^ 5 = to 15; 15 to 30; 30 to 40; 40 to 45; 45 to 50; 50 to 60,. 
Make (7= 4000; 3500; 3000; 2500; 2000; 1500. 

103. The Pencoyd Tests of Full-size Rolled-iron I-beams, 
Channels, Angles, Tees, etc., used as Columns. — These were made 
in 1883 by Mr. Christie, at the Pencoyd Iron Works of Phila- 
delphia, Pa., and were very careful and extensive. The following 
table is based on them, ^y '^ fixed ends " it is here meant that 
the ends are so securely attached to the contiguous supports that 
the fastenings would not be ruptured if the column were sub- 
jected to a breaking load; by "flat ends," that the ends are 
squared off and bear on a flat, firm surface. " Hinged ends " 
refers to the ends being fitted with pins, or ball-and-socket joints, 
of proper size, with centres practically in the axis of the column 
(this axis being the line containing the centres of gravity of all 
sections of the column, which is prismatic); while ''round ends" 
implies that the ends have only points of contact such as balls or 
pins bearing on a flat plate, the point of contact being in the 
axis of the column. The first column of the table contains the 
ratio of the length, U of the column to ^, the least radius of gyra- 
tion of the section (except that for hinged ends the pin must be 
at right angles to the least radius of gyration). Factors of safety 
are recommended as given, being different for flxed and flat ends 



THE PENCOYD EXPERIMENTS WITH COLUMNS. 



113 



Ratio 














I 
k 


Fixed Ends. 


Flat Ends. 


Factor of 
Safety. 


Hinged Ends. 


Round Ends. 


Factor of 

Safety. 


20 


46,000 


46,000 


3.2 


46,000 


44,000 


3.3 


40 


40,000 


40,000 


3.4 


40,000 


36,500 


3.6 


60 


36,000 


36,000 


3.6 


36,000 


30,500 


3.9 


80 


32,000 


32,000 


3.8 


31,500 


25,000 


4.2 


100 


30,000 


29,800 


4.0 


28,000 


20,500 


4.5 


120 


28,000 


26,300 


4.2 


24,300 


16,500 


4.8 


140 


25,500 


23,500 


4.4 


21,000 


12,800 


5.1 


160 


23,000 


20,000 


4.6 


16,500 


9,500 


5.4 


180 


20,000 


16,800 


4.8 


12,800 


7,500 


5.7 


200 


17,500 


14,500 


5.0 


10.800 


6,000 


6.0 


220 


15,000 


12,700 


5.2 


8,800 


5,000 


6.3 


240 


13,000 


11,200 


5.4 


7,500 


4,300 


6.6 


260 


11,000 


9,800 


5.6 


6,500 


3,800 


6.9 


280 


10,000 


8,500 


5.8 


5,700 


3,200 


7.2 


300 


9,000 


7,200 


6.0 


5,000 


2,800 


7.5 


320 


8,000 


6,000 


6.2 


4,500 


2,500 


7.8 


340 


7,000 


5,100 


6.4 


4,000 


2,100 


8.1 


360 


6,500 


4,300 


6.6 


3,500 


1,900 


8.4 


380 


5.800 


3,500 


6.8 


3,000 


1,700 


8.7 


400 


5,200 


3,000 


7.0 


2,500 


1,500 


9.0 


430 


4,800 


2,500 





2,300 


1,300 




440 


4.300 


2,200 





2,100 







460 


3,800 


2,000 




1,900 







480 




1,900 





1,800 








from those proposed for hinged and round ends. In the other 
columns the numbers given are the respective breaking stresses 
in lbs. per sq. in. of sectional area, v^^hich number must be multi- 
plied by that area in sq. in., for the actual breaking load ; divid- 
ing which bj the ])roper factor of safety we obtain the safe load 
in pounds. 

I^'or example, required the breaking load of a 9-in. light iron 
I-beam of the IS". J. Steel and Iron Co., 14 ft. long and used as a 
column with flat ends. 

From p. 40 of the hand-book we find: least /= 4.92 and 
the area of section = 7 sq. in. Hence the least ^' is I ~ F, 
— 0.703, and h itself = 0.838 in. ; so that Z -^ A; = 168 in. -^ 
0.838, = 200. The breaking load, then, is FC ^ 7.00 X 14,500 
~ 101,500 lbs. ; and the safe load would be 101,500 ^ 5 = 20,300 
lbs. (If Eankine's formula were apphed to this same case a fair 
agreement with this result would be found.) 



114 



E"OTES Al^D EXAMPLES l^N MECHANICS. 



^t-> 



:iY 



104. Cast-iron Channel as a Column. — In the case of the 
channel-shaped section in Fig. 125, com- 
posed of three rectangles, all of the same 
width, =z t, it is desirable, for economy of 
material if the prismatic body is to be 
used as a column, that the moments of 
inertia about the two gravity axes ^ and 
1^ should be equal (see § 311, M. of E.). 
The problem, therefore, presents itself in 
this form : Given the width h of the base 
DC of the section, and the thickness of 
metal t, what value should be given to the 
length G, of the projecting sides AD and 
BC, of the channel to secure this result? 

The algebraic statement of the conditions involved leads to 
an equation of high degree for the unknown quantity c. But, 
hy numerical trial, a few reliable values have been found, given 
below, by simple interpolation between which all ordinary cases 
in practice may be satisfied. 



Fig. 125. 



When ^ = (infinitely thin), 



t = O.OSSb = -^h, 
t = 0.166b 



i^ 



make c = l.S7h 
" c = 1Mb 
" = 1.25b 



t = 0.500b = U 



\ solid ), 
( rectanorle ) 



1.00b, 



In the construction of such a column the edges A and D of 
the projecting sides should be tied or braced together at inter- 
vals ; or occasional transverse webs may be introduced. 

It is remarkable in this problem that the distance u of the 
centre of gravity (7 from the base DO is almost exactly equal to 
one half of h in every instance, and may be so assumed in locat- 
ing the axis of the column that the load may be applied in that 
axis. 

105. Vertical Reactions of Horizontal-faced Piers bearing a 
Beam and Loads. Graphical Method. — The construction on p. 404 
of M. of E. is the one usually given and is the most convenient; 
still, the proof is a little puzzling to the student because the 



PIER-REACTTO]SrS — GRAPHICS. 



115 



amounts and direction of the two auxiliary, mutually annulling, 
forces P and P' are not known at the outset. 

Hence we here present a construction in which those two 
forces are completely assumed and known at the outset. Assume 
1 and 6 as the auxiliary forces (equal, opposite, and in the same 
line), there being three loads in this instance, 2, 3, and 4; Fig. 
126c J^umber the forces of the system as in figure and draw a 



1- \ y\ 




Fig. 12i 



portion of the force-polygon with the known forces 1, 2, 3, and 
4, beginning at 0^ and the first three rays, I, II, and III (dotted). 
The first segment of the corresponding equilibrium polygon 
should begin at a^ the intersection of the action-lines of forces 1 
and 2, and finally the third segment cuts the action-line of 5 in 
the point m. JS'ow the fourth segment is the last in this case [of 
three loads, 2, 3, 4], and must pass through the intersection of 
the last two forces, 6 and 7, i.e., through A. Hence draw inA 
and a parallel to it tlirough the pole 0^ this latter line being the 
fourth ray desired, whose intersection with the "load-line" at n! 
cuts off the proper length of the right-hand reaction 5. The 
forces 6 and 7 are now easily added to the force-diagram in an 
obvious manner and the latter is complete ; the magnitude of the 
other reaction, 7, being thus determined. Of course, the value 
of force 7 is also given by the n'd^ and the force-polygon could 
also be closed by running from n' to d and then from d to 0. 
instead of in the manner shown. 

IS'ote the order of numbering in the above. The two assumed 
forces are made the first and the last hut one^ respectively ; the 
unknown reactions are made the last hut two^ and the last^ 
respectively ; while the given loads are assigned to the interven- 
ing numbers, in any order. 



116 



NOTES AND EXAMPLES IN MECHANICS. 



K — «• — 4H 



v/x 



Tr[Tn>s^ ,„ 




106. Construction for Use with the Treatment in § 390, M. of 
E. (Graphical Statics). — For the particular case involved in the 
treatment of the straight horizontal girder, built in, of § 390, 
M. of E., the following is given, to replace the more general 
construction of § 377, M. of E. 

At (I.) in Fig. 127, we have a curve, or broken line (equi- 

, librium polygon, for instance), 

I FKWMG^ connecting points F 

"^l and G in two vertical lines. 

Across this curve we wish to draw 

I a right line v . . m, in such a way 

that the area K WM above v . ,m 

i\ shall be equal to the sum of the 

areas vFK and mGM below 

I 'y . . m, and also that the centre 

I of gravity of the upper area shall 

be in the same vertical line as 

that of the two lower, or negative 

areas, taken together. 

Only one position of v , , m 
will do this, the algebraic expression for which is that 2(2') = 0, 
and that 2(xb^) = ; (the areas in question being divided into 
vertical strips of equal horizontal widths = /^x, the distance of 
any strip from the vertical line vFhemg called x.) 

Annex the figure FGMK (having joined F , . G) to both 
the positive and negative figures above mentioned and the con- 
dition now becomes [see (11.) in figure] that the area of the 
figure FKWMG , . F (the right line FG being its lower 
boundary) must equal that of the two triangles vFG and niGv, 
and that the centre of gravity of the former must lie in the same 
vertical as that of the two triangles combined. In other words, 
if the area of the curvilinear figure, FKWMG-F, with FG as 
base, be considered as a weight R acting through its centre of 
gravity, then the areas of the two triangles must represent the 
two upward reactions /'and T' of two piers [see (III.) in figure] 
supporting a horizontal beam on which R rests. These imagi- 
nary piers are evidently at distances of one third the span from 



....^^FEE^ri. 



Fig. 127. 



EQUILIBRIUM POLYGON- THROUGH THREE P0I:N^TS. 117 

the verticals through F and G. Adopt, therefore, the following 
construction : 

By dividing into vertical strips find the area of the curvi- 
linear figure with base FG. Draw the pier verticals at the one- 
third points. Find the vertical containing the centre of gravity 
of the curvilinear figure by p. 415. Compute or construct the 
values of Tand T' on the conception of the known area R being 
a weight supported on the beam with T and T' as reactions. 
{T and T' are most easily obtained, perhaps, by scaling the 
distances s and t (see figure) and writing T{8 -\-i) ^i Rt, and 
T' = R- T.) 

If the area of the triangle FGv must be = T, one of the 
values just found, knowing that this area = -J (altitude I) X base 
vF, the proper length of vF is easily computed ; and similarly, 
using T' and the triangle mvG^ we calculate 7nG. Joining v and 
m, the required right line vm is determined. 

107. Three-point Construction. Equilibrium Polygon for Non- 
vertical Forces. Preliminary Step. — In the construction and proof 
of § 378a, M. of E., it is supposed at the outset that the action- 
line of the resultant (i?i) of all the forces acting between points 
A and ;p has been found ; similarly, that of the resultant {R^ of 
the forces acting between p and B ; and that of the resultant {R) 
of the two partial resultants. 

It is here intended to give the detail of finding these three 
lines and to make clearer the scope and intent of the problem. 
Fig. 128. Let A, p^ and B be the three points through which 
the equilibrium polygon is to pass, and 1, 2, etc. (to 6 inclusive) 
(on left of figure), the given forces in magnitude and position ; 
1, 2, and 3 acting between A and p ; and 4. 5, and 6, between p 
and B. Lay off the " load-line" ST&\ on some convenient scale of 
force, and select any pole as 0^'\ Draw the "rays," O'^^S, etc. ; 
and also lines parallel to them, in proper order, beginning at a 
convenient point JL^^ (not Jl, necessarily), so as to form an equilib- 
rium polygon, A"B", as shown. The first segment is parallel 
to 0"'S\ the last, to 0'" U \ and the segment connecting forces 
3 and 4 (in this particular instance), parallel to 0"'T. The inter- 
section of the first and last segments gives M\ a point in the 



118 



NOTES AND EXAMPLES IN MECHANICS. 



action-line of R, the resultant of all the given forces, 1 . . 6 ; and 
similarlj, the other intersections N" and 0" are points in the 
action-lines of R^ and R^^ respectively. Right lines through 
these three points, parallel respectively to 8U^ ST^ and TJJf 



Preliminary step; Equil., 
Polygon thro' Three Points • •' 
Porces not Vertical. '/ 


m" 

R \ 


/ 

A 
/ 


# 


N>5< V ! h / y 

A'--\ 1/ 

,/ ©A " \ !/ 


r/, 


Bo^ 


w 



Fig. 128. 

should meet in a common point 6", and are the respective action- 
lines desired. 

The further steps are those given on pp. 460 and 461, M. of 
E. Our present equilibrium polygon, A" B'\ is of no further 
use ; but the load-line STU will still serve, after the lines M' A 
and M' B have been located according to § 378a. We can then 
draw through S and TJ parallels to M' A and M B^ respectively, 
and by the intersection of these parallels with each other deter- 
mine the pole corresponding to the final equilibrium polygon 
which is to pass through A^jp^ and B. 

It is to be noted that the problem of the '' Shear-legs" of § 59 
of these Notes, and also Problem 2 (of the two links), p. 35, M. 
of E., are cases of a three-hinged arch-rib, and can be treated 
graphically in the same manner; and thus the "special" equilib- 
rium polygon and its corresponding pole and rays (i.e., force- 
diagram) determinedc 

The ray parallel to the segment passing through the inter- 
mediate joint (^) gives the amount and direction of the pressure 
on the hinge of that joint ; and corresponding statements may be 
made for the two extreme joints. 



CHAPTER YIII. 

Miscellaneous Notes. 

108. Co-ordinates of Centre of Gravity — Fuller Explanation. — 
Assume the various small particles of a rigid body to be num- 
bered 1, 2, 3, etc., and call their respective volumes dV^, dV^, 
d F3 , etc. (cubic feet), and their a?-co-ordinates x,, x^,x^, etc. 
(feet). If the body is heterogeneous, the particles may be of dif- 
ferent densities; for example, particle 1 may be of such a density 
that a cubic foot of material of that density would weigh 100 lbs., 
while a cubic foot of the material of which particle 2 is composed 
would weigh 110 lbs. ; and so on. Or, in symbols, the " heavi- 
ness," or rate of weight, of particle 1 is ;^, = 100 lbs. per cubic 
foot, while that of particle 2 is ;;/2 = 110 lbs. per cubic foot. A 
similar notation would apply to all the other particles. 

The respective weights, then, of the particles (or force of the 
earth's attraction on them) are y^dY^^ y^dV^^ y/lV^^ etc. 
(pounds), and if we substitute these for the forces Pj , P^^ P^ 
etc., in the expression for the x'oi the centre of parallel forces 
(foot of p. 16, M. of E.), we obtain 

r/lV, + y,dK + y,dV,+ ... '• • • W 
which in the compact notation of calculus, the particles being 
taken " iniinitely small " and, therefore, " infinite " in number, 
can be written 

fxydV _ ^ 

x=-z . or x^-^JxydV, . . . . (b) 

/ ydV 

where G is the total weight of the body, = I ydV, 

If the body is homogeneous^ all the particles have a common 
" heaviness," which we may call y,^ and factor out, thus obtaining 

119 



120 NOTES AND EXAMPLES IN MECHANICS. 



""- r.n[dv, + dr,+ ...] ' ^" ^- 

^^rom which the y„(iau be cancelled, leaving 



r, 



r«fdr 



\rf«:dV, (c) 



where F denotes the total volume of the bodj. (ISTote that the 
factoring out of a common multiplier from a parenthesis corre- 
sponds to taking a constant outside of the integral sign.) 

109. The Time-velocity Curve and its Use. — From eq. (I.), p. 50, 
M. of E., we have ds -^ dt ^:^ v, the velocity of a moving point 
at any instant. Hence, also, ds = "vdt, or the element of dis- 
tance equals the product of the velocity at that instant by the 
element of time. If now, whatever the character of the recti- 
linear motion, we conceive a curve to be plotted, in which the 
time (from some initial instant) is laid off as an abscissa, and 
the velocity of the moving point as an ordinate, this curve 
may be called a " time-velocity " curve for the particular kind of 
motion, being different for different kinds of motion. (Of 
course, proper scales must be selected in laying off distances on 
the paper to represent the quantities time and velocity.) 

The equation expressing the relation between the two vari- 
ables, time and velocity, may be regarded as the equation to the 
curve. Thus, in uniform motion we have the velocity v con- 
stant, so that the curve is a right line parallel to the horizontal 
axis, or axis of time, as AL in Fig. 130, where OA represents the 
constant velocity. If the motion is uniformly accelerated^ that 
is, has a constant acceleration, we have v ■=^ v^ -\- V^ [where p is 
the acceleration^ or rate of change of the velocity, and v^ is the 
initial velocity (for z5 = 0)], and note that the quantity ^z5, or 
total gain in velocity over the initial velocity v^^ is directly pro- 
portional to the time, so that the curve obtained is a straight line 
inclined to the axis of abscissas, as BII in Fig. 130, BO repre- 
senting the initial velocity v^ . 

In general, let KYM^ Fig. 129, be the time-velocity curve 



MISCELLANEOUS NOTES. 



121 



for any rectilinear motion. Here we note that the product v . dt 
at any instant during the motion is represented by the area of the 
vertical strip YR^ whose width is dt and length v (by scale). 
Hence the element of distance ds (feet) is proportional to that 
area, and the whole distance {s) described from the beginning is 
represented by the sum of the areas of all such strips from OK 
to VR^ i.e., by the area OKYR\ while the distance (^„) de- 
scribed between t =^ ^ and t = tn will be represented by the 
complete area OKYMN, 

If now ON be regarded as the base of the figure OKMN^ we 
may conceive of a rectangular figure OALN having the same 
area and same base as OKYMN, and having 
some altitude -A 6^ which can be looked upon 
as representing the " average velocity " of the 
motion between ^ = and t = t^. By " aver- 
age velocity " would be meant that constant 
velocity necessary in a uniform, motion to 
enable a moving point to describe the dis- 
tance Sn in the same time t^ as in the ac- 
tual motion. In other words, the "aver- 
age velocity " is the result obtained by di- ^^^- ^^^• 
viding the whole distance by the whole time, and is represented 
by the altitude AO^ since the area of a rectangle is equal to the 
product of its base, 0N{~ t^), by its altitude, AO. 

As a useful instance consider again a uniformly acceler- 
ated motion. Its time- velocity curve is a straight line, BH^ 
Fig. 130, the initial velocity v^ being rep- 
resented by OB, and the final, v^ (for 
t z= tn), by UN The distance described is 
represented by the area of the trapezoid 
OBHN, This area is equal to that of a 
rectangle of the same base ON (i.e., t^ 

^ ^„. and of an altitude OA = half the sum of 

Fig. 130. 

OB and NB; i.e., s^ = i{v^ + Vn)tn- In 

other words, the average velocity between t = and t = t^ for 

the uniformly accelerated motion is ^{v^ -\- v^. 





122 NOTES AND EXAMPLES IN MECHANICS. 

If the initial velocity of the uniformly accelerated motion is 
zero, the time-velocity curve becomes a straight line passing 
/ through the origin, 0^ viz., OH^ Fig. 131 ; 
/ and in that case the distance ,9^ is repre- 
sented by the area of the triangle OHN^ 
and the average velocity OA is one half the 
final, or the final is double the average. 
Therefore to obtain the whole distance de- 
scribed we must multiply the whole time by 

^ < t-n > one half the final velocity ; i.e., s^ = ^J^n- 

Fig. 131. jf^ then, in this case of uniformly ac- 

celerated motion with initial velocity = zero, we divide the whole 
distance by the whole time, we do not obtain the final velocity (a 
common error with students), but only the " average velocity," 
in the sense defined above. This result must then be doubled 
to obtain the final velocity. 

As an instance where the average velocity is one third of the 
final (the initial velocity being zero), consider the case of variably 
accelerated motion represented by the relation v = qf, where 
^ is a constant. The time-velocity curve will have the form 
OFB, Fig. 131, a parabola with vertex at 0. The area OPEN 
will be one third of that of the circumscribing rectangle, and 
hence is equal to 'ON X one third of i?S, i.e., to ON X OB. 
Hence OB^ or one third of the final velocity, is the average 
velocity. 

Or, mathematically, in detail, ds^ = vdt, = qfdt ; 
therefore 

Sn = qJ^'ydt = iqf^' = [iqtny^. . . . (1) 
But for t = tnwe have v, = v^, = qfn, and hence 

Sn = iiVnYn • (2) 

That is, the average velocity is equal to one third of the final. 

110. Reduction-Formulae for Moment of Inertia of a Plane 
Figure. — (To replace § 88, M. of E., as regards the Mo7nent of 
Inertia of a plane figure.) 



MISCELLANEOUS NOTES. 123 

Definition, — Any right line containing the centre of gravity 
of a plane figure is called a " gravity axis " of that figure. 

Theorem. — The moment of inertia of a jplane figure about 
a given axis in its own plane is egual to its moment of inertia 
dbont a gravity axis parallel to the given ___^— ^\ 

axis, augmented hy the prodicct of the area ^^ q^f" / 

of the figure hy the square of the distance f \ \ ) . 

between the two axes, Q-j 

Proof.— F\g. 132. Let dF be the area ( 
of any element of the plane figure, and z' v. 
the distance of that element from any axis 
X in the plane of the figure ; while z is its | j 

distance from a "gravity axis," g, parallel ~ 

to axis X. By definition we have 7^ == ^^^' ^^'^^ 

iz'^dF\ but z' = z -{- d, d being the distance between the two 
axes. Hence 

jT = ^(3 _|_ d)'dF= Jz'dF-^ ufzdF+ d'fdF, 

1^0 w, from the theory of the centre of gravity, we have (see 
eq. (4), p. 19, M. of Y.^ fzdF — Fz, where z is the distance of 
the centre of gravity C from the axis g. But ^ is a gravity axis, 
so that ;3 = ; and hence fzdF — 0. Also fdF — F^ the whole 
area ; and fz^dF — Ig. Whence, finally, 

I,:=Ig + Fd\ . . . (4) Q. E. D. 
It also follows, by transposition, that 

Ig^I^-Fd"', (4a) 

which shows that the moment of inertia of a plane figure about 
a gravity axis is smaller than that about any other axis parallel 
to that gravity axis. 

The moment of inertia of a plane figure plays an important 
part in the theory of beams subjected to bending action, the 
transverse section of the beam forming the plane figure in ques- 
tion ; somewhat as the mere area of the section does when the 
beam or rod is subjected to a straight pull. 



124 NOTES AND EXAMPLES IN MECHANICS. 

111. Miscellaneous Examples. (See opposite pages for figures.) 

i. Fig. A. Given the data of the figure, find the stress in every two-force 
piece, and the three pressures exerted on the pin at B. 

2. Wind from the S.W,, 30 miles per hour. Ship going toward the N.W. 
at 10 miles per hour. At what angle with ship's course should a vertical sail 
be placed that the air-particles may strike it at an angle of 30° on the hinder 
side ? 

3. Locate, by calculus, the centre of gravit}"- of the plane figure in Fig. B, 
the equation to the upper bounding curve being xy = 20 sq. ft. 

4. In the rectilinear motion of a material point weighing 12 lbs., and mov- 
ing horizontally on a rough surface, friction from which is the only horizon- 
tal force, we note that positions A, B, C, and D are passed at the following 
times (by the clock), respectively : 

3 h. 4 m. 8.1239 sec; 3 h. 4 m. 8.2350 sec; 3 h. 4 m. 8.3490 sec; 
3 h. 4 m. 8.4658 sec 

The distance from AioB\& 10.00 ft. ; from B to C, 10.02 ft. ; and C to D, 
10.05 ft. 

Find (approx'ly) the acceleration of the motion as the material point 
passes the position B ; also for G; and what must be the value, in pounds, of 
the friction at 5 ? 

5. Fig. C. Fiud the stress in each two-force piece, and pressure on pin at 
A ; also pressure of pin B on bar BD. 

6. Of the solid right cylinder in Fig. D, the internal conical portion is of 
lead, whose spec. grav. is 1L3, while the remainder is of cast iron. Find the 
centre of gravity of the whole solid. 

7. In Fig. 80 on p. 67, Notes, compute the pressure between the block and 
the guide when the former, having started from rest at A, is passing position 
B, 45° from A. 

8. Fig. E. The block weighs 40 lbs. The cord d . . . e is, attached to it 
at angle 15° with the plane, which is smooth. If the block is to be permitted 
to slide from rest down the plane, under action of gravity, the cord, and 
the plane, what constant tension (lbs.) must be maintained in the cord that a 
velocity of 12 ft. per sec. may be generated in 2 seconds ? Afterwards, what 
new value must be given to this tension if the velocity is to continue at that 
figure (12 ft. per sec.) ? 

9. In the vertical fall of a material point, certain consecutive small space- 
intervals are given, = a, b, c, and d ; and the corresponding time-intervals, 
t, t', t", t'" . Derive approximate formulae for the velocities at mid-poiuts of 
these time-intervals ; the accelerations near the end of some of these intervals ; 
and the corresponding resultant force that must be acting, the weight of the 
body being O. 

10. Fig. F. Find the amount and position of the pressure between the bar 
AB and each of the four pins passing through it ; also the stress in BB. 




400 lbs. 



B 





1200 lbs. 



1^-6^- 






/ ^ *. 





0/15°/ / 

J-y 


"4 


wk/ 


7 


40° 

1 





20i6s. 



23Z&S. 






[A is inside 
o/B] 





h< 5- *+ 



JV.S. Thesmallpulleyisfree 
on the axle of wheel B . 




2400 ibs. 





20' 





—friction - 
12000 lbs. 



J 



126 NOTES AND EXAMPLES IN MECHANICS. 

11. Fig. G. The two pulleys run on fixed bearings, without friction. If 
the given three weights be allowed to come to a position of equilibrium, the 
knot A being a fixed knot, find the angles the two oblique cords make with the 
vertical, respectivel}^ 

12. Find the moment of inertia about the axis X of the symmetrical 
plane figure shown in Fig. H. The equation to the bounding curve AB is 
{x — a') -^{a — a') = z"^ -^ Ic^. 

13. Fig. J. The block weighs 13 lbs., and in sliding down the rough in- 
clined plane encounters a variable friction, which in lbs. = 3 times the dis- 
tance, s, from the starting point, in feet. It starts from rest at A. Find the 
velocity acquired on its reaching a position 3 ft. vertically below A. 

14. Compute the moment of inertia of the plane figure in Fig. K about the 
^ axis X; also the corresponding radius of gyration. 

15. Compute the y co-ord. of the centre of gravity of the plane figure 
shown in Fig. L. The equation to bounding curve is x^ =4^, with the foot 
as linear unit. 

16. A block of 50 lbs. weight is started along a rough horizontal table with 
an (initial) velocity of 40 ft. per sec. If the friction met with is variable, and, 
in lbs., equal to 700 times the body's weight -^ (1300 + the square of the 
veloc. in ft. per sec), find the time and distance in which the body comes to 
rest. 

17. Fig. M. Find the tensions in all chains and the pressure on pin A and 
under wheel B. 

18. Fig. N. Find the moment of inertia of the plane figure about the grav- 
Jty-axis X which is perpendicular to the axis of symmetry. 

19. Fig. P. The ram A, of 1300 lbs. weight, falls from rest through 30 
ft., and has then an inelastic impact with the pile B, of weight 400 lbs. Com- 
pute their common velocity after the impact, the Kinetic Energy lost in the 
impact, and the distance the two bodies will sink after the impact, overcom- 
ing the constant frictional resistance of 13000 lbs. on side of hole. 

30. Fig. Q. Block of 10 lbs. weight on inclined plane. It starts from rest 
at A. If friction on AB is 3 lbs., while on BG it is 3 lbs., compute the time 
of reaching position G, and its velocity at that instant. 

31. Fig. R. A block of 13 lbs. weight falls from rest, freely through the 
first 7 ft., but then strikes the head of spring A, which opposes a resisting 
force at rate of 100 lbs. per inch of shortening ; and 3 inches further down 
strikes spring B, offering 160 lbs. of resistance per inch of shortening. Where 
is the block when (momentarily) brought to rest (supposing the elastic limit of 
the springs not passed) ? 

33. A bod}' weighing 13 lbs. is given an initial upward vertical velocity of 
10 ft. per sec, being thereafter acted on only by gravity and a variable hori- 
zontal force = {yq ^^ ^^^^ ^^ •iQc'] lbs. Find the equation to its path. 






B 










H 



a — » ,---B 




20 its. 



A 
10 lbs. 



22 lbs. 






[A Js inside 
o/B] 





._5i' vj- 



N.B. The small pulley is free 
on the axle of wheel B . 




2m) lbs. 




o^-'-- 





2b' 




j^^ -friction ^ 

mm lbs. 

I 



128 NOTES AND EXAMPLES IN MECHANICS. 

112. Answers to Preceding Problems.— 1. 339.1 lbs. and 240 
lbs. ; on pin A, 240 lbs. vertically, 288.3 lbs. at 33° 42' with hor- 
izontal, and 466.2 lbs. at 30° 56' witli vertical. 

2. 4i° 34' ; or sail pointing 3° 26' N. of West. 

3. X = 3.28 ft. ; y z^ 3.28 ft. ; area of Hgure = 18.31 sq. ft. 

4. In passing B, ace. = 18.72 ft. per sec. per sec. ; C, 16.05; 
friction as the material point passes £, 6.98 lbs. 

5. 154 lbs. compression, and 1200 lbs. tension ; at A, pressure 
= 738 lbs. at 25° with horiz. ; pressure of pin £ on bar BD = 
1132 lbs. at 36° 45' with vertical. 

6. X = 5.39 inches from left-hand base. 

7. Pressure is 42.44 lbs. 

8. First value of tension, 18.92 lbs. ; second, 26.63 lbs. 

9. ISTear the end of the first interval the acceleration = 

Tit'+^'f) ' *'^"* "'^^ ""^ "^ ''''"'^ '""'''"'^^ S(7T?)- 

10. Pressures are 1600, 457, 2263, 1143 lbs., respectively. 
There is no stress in piece DB. 

11. On the left, 87° 43' ; on the right. 65° 17'. 

12. The moment of inertia, about X, is f[ah^ + ^a7i ^]. 

13. The velocity acquired is 7.13 ft. per second. 

14. /^ =: 330.66 biquad. inches; rad. gyr. = 3.02 inches. 

15. The area is 80 sq. ft. and y = 15.61 it. 

16. Time = 3.075 sec, to come to rest. Distance =70.9 ft. 

17. Tension in upper chain, 5838 lbs. ; in the lower oblique 
chain, 2770 lbs. Pressure under wheel B, 1385 lbs. Pressure 
on pin at A is 7503 lbs., and is directed toward the right at an 
anoxic of 6° 37' above the horizontal. 

18. Distance of centre of gravity from the horizontal line 
drawn through the upper corners is 2.20 inches. Moment of 
inertia about the gravity axis JT is 8.12 biquad. inches. 

19. 26.91 ft. per sec. ; 6000 ft.-lbs. lost ; 1.73 ft. sinking after 
impact. 

20. Time, A to (7, = 2.06 sec; velocity at 6^=18.05 ft. per sec. 

21. The block is 7 ft. 4.297 inches below its starting-point, 
i.e., has shortened spring A an amount of 4.297 inches. 



PROBLEMS AND EXAMPLES. 



129 



Q 



\< 1-'- 



2) B 



(E 



U— 



— I 

Fig. H9. 



W=wZ 



^ 



i" 



Fig. 150. 



113. Examples. — 1. Fig. 149. The hollow shaft A is to be 
twice as strong torsionallj (i. e. , as to tor- 
sional moment) as the solid shaft B ; while 
the material of A. is only half as strong as 
that of B. Given the lengths I and l'\ 
the radius r'^ of B, and the outer radius, 
r, oi A, determine a proper value for r\ the inner radius of A, 
for above conditions. 

2. Fig. 150. Homogeneous prismatic beam; rectangular 
section ; width = h smd height = h ; placed with h horizontal, 

and loaded uniformly over its whole 
length at rate of w lbs. per running 
inch. Given the whole length, I, 
-H and position of support A (at left 
1 end), where (i.e., distance a =^ ?) 
shall we place support jB that the 
moment of stress-couple in section over B shall equal (without 
regard to sign) the greatest moment of stress- couple occurring 
between A and the point of inflection, i^, of the elastic curve ? 

Also, after a is found, find the maximum shear, (h and h 
are given, and also w ; elastic limit supposed not passed.) 

3. Fig. 151. By the principles of the graphical statics of 
Tnechanisni, having the resisting force Q given in 
amount and position, given also all friction angles con- 
cerned, and considering all kinds of frictional action, 
find the value of P for forward motion ; also for back- 
ward motion ; the 
efficiency of the 
mechanism. 

4. Fig. 152. The 
load, of weight P 



■=z 60 lbs., is 



gradu- 





Fig. 151. 



Fig. 152. 



ally applied at lower 
end of compound vertical round wire (of wrought iron) whose 
upper end is fixed. Neglecting the weight of the wire, compute 
the total elongation of the wire and the work done on the wire 



130 



NOTES AND EXAMPLES IN MECHANICS. 



during tlie gradual stretching. (Lengths are 100' and 120'; 
diameters J in. and Jg- in., respectively.) 

5. Fig. 153. The horizontal prismatic beam is of rectangular 
section ; widtli J, horizontal. Its height A is to be three times 
its width. Beam of timber. The length and character and 
amount of loading are given in the figure, the weight of the 
beam itself being neglected. 

There being three (local) maxi- 
mum moments (of stress-couple), 
viz., at A^ B^ and (7, locate sec- 



5 tons 



1 ton 



w 

it]- 



I 

— 10-- 



FiG. 153. 



itto-n^ tion B bj determining distance 
" « " ; compute the moments at 
A^ B^ and C\ and then determine the minimum safe dimen- 
sions to be given to the section of the beam. (The load of ^yq 
tons is uniformly distributed along the ten feet.) 

6. Fig. 154. The short vertical cylindrical body abed is 
fixed at the upj^er end while sustaining at its lower 
end a weight of 4000 lbs., whose line of action 
prolonged upward passes through the extreme 
edge, (3^, of the section ah. 

The horizontal section of the body is a circle 
of radius =0.5 inch. It is required to compute 
the stress per square inch at point a of the section 
db.^ and also that at point b. 

Also, what would these stresses be if the line 
of action of the 4000 lbs. load passed- through the 
centre of the section ab ? 

114. Answers to Problems in §113. (For 6, see below. ■^) 

1. t' = \^r* ^rr"^. 

2. a=(l— Vi)l, = 0.293^. The maximum shear is just 
on the left of support B and is e/^ = ( ^2 — l)wl = 0.4:14:wl. 

3. (Solved graphically.) 

4. Total elongation =1.21 inches. Work done = 36.3 
in. -lbs. 

5. Distance a = 4:A ft. Moments are 1.00, 3.84, and 4.00 

♦.Answer to Ex. 6 : — At a, 12.7 tons per sq. in. tension ; at b, 7.63 tons per 
sq. in. compression. 




IN EOTARY MOTIO]^. 131 

ft. -tons, respectively. Taking R' = 1000 lbs. per sqo in., we 
have A = 12, and 5=4, in. 

115. The "Imaginary System.'^ — In conceiving of the imagi- 
nary equivalent system in § 108, M. of E., applied to the 
material points supposed destitute of mutual action, and not 
exposed to gravitation, we employ the simplest system of forces 
that is capable, by the Mechanics of a Material Point, of pro- 
ducing the motion which the particles actually have. If now the 
mutual actions, coherence, etc., were suddenly re-established, 
there would evidently be no change in the motion of the assem- 
blage of particles ; that is, in what is now a rigid body again ; 
hence the imaginary system is equivalent to the actual system. 

In applying this logic to the motion of translation of a rigid 
body (see § 109 and Fig. 122, M. of E.) we reason as follows: 

If the particles or elementary masses did not cohere together, 
being altogether without mutual action and not subjected to grav- 
itation, their actual rectilinear motion in parallel lines, each hav- 
ing at a given instant the same velocity and also the smne accel- 
eration^ j>^ as any other, could be maintained only by the appli- 
cation, to each particle, of a force having a value ^ its mass X p^ 
directed in the line of motion. In this way system (II.) is con- 
ceived to be formed and is evidently composed of parallel forces all 
pointing one way, whose resultant must be equal to their sum, viz. 

/ dM X P' But since at this instant p is common to the motion 

of all the particles, this sum can be written p I dM, = the whole 

mass M X p- 

If now the mutual coherence of contiguous particles were sud- 
denly to be restored, system (II.) still acting, the motion of the 
assemblage of particles would not he affected (precisely as the fall- 
ing motion in vacuo of two wooden blocks in contact is just the 
same whether they are glued together or not) and consequently we 
argue that the imaginary system (II.), is the equiv&.lent of what- 
ever system of forces the body is actually subjected to, viz. sys- 
tem (I.), (in which the body's own weight belongs) producing the 
actual motion. 




132 NOTES AND EXAMPLES IN MECHANICS. 

Since the resultant of system (II.) is a single force, = Mp^ 
parallel to the direction of the acceleration, it follows that the 
resultant of the actual system is the same. 

116. Angular auantities— Rotary Motion about a Fixed Axis. 
—In Fig. 155 let the body J/ZiV^ rotate about the fixed axis G 
(perpendicular to paper), the initial posi- 
tion of the arm CL having been CY. 
Now the angular motion of the whole 
rigid body is the same as that of the arm 
CL. If in a small time- interval, dt^ the 
arm passes from position CA to position 
CB^ and thus describes the small angle 
ACB^ whose value in ;r-measure is da 
radians, the angular velocity at about the mid-point of angle 
A GB is GD=^ da-^ dt. In the next and equal time-interval a 
slightly different angle, da' radians, is described ; and if in the 
figure we lay off angle ^(7^ equal to AGB^ the angle EGB^ or 
difference between da and da may be called d'^a. And so on, 
for any number of consecutive ((^6if)'s, described in equal times, 
each = dt. If the motion is uniform^ all the {day^ are equal ; 
that is, the angular velocity is constant, each d^'a being = zero. 

If all the ((^^«f)'s are equal, the motion is uniformly accel- 
erated^ the angular acceleration, ^, being thus determined : The 
gain, or change, of angular velocity occurring between the mid- 
point of AGB and that of BGD is — —^ or go' — gd\ and 

Cut at 

if this be divided by the time, dt^ occupied in acquiring the 
gain, we have for the rate of change of angular velocity, that 

GO — GO d Gl) 

is, for the angular acceleration, the value 6 — — -r. — = -rr. 

Another form is this: if the gain of angular velocity, 

— , be divided by dt, we have 

dt dt' -^ ' 

__ da' - da ^ d'a ^ ^ 

{dty df ^ ^ 

If the successive {d'^ays are unequal (the successive {days 



ANGULAR QUANTITIES. ROTARY MOTION. 133 

being described in eqvjdl time-intervals, it must be remembered), 
the motion is some kind of variahly accelerated angular motion ; 
e.g., in a harmonic (oscillatory) rotary motion the angular 
acceleration at any instant is directly proportional to the angular 
space between the ' ' arm ' ' or reference line of the body and the 
middle of its oscillation, and is of contrary sign; i.e., 6= —Aa, 
where ^ is a constant. 

JExample 1. At a certain part of its revolution a fly-wheel 
is found to describe just 1° in 0.01 second. Here we have 
da:=z 0.01745 radian, and dividing this by the dt,= 0.01 sec, 
we obtain go = 1.745 radians per second as the angular velocity 
of the wheel at this part of its progress, as nearly as the data 
permit. (Strictly, this value of go is only the average value of 
the angular velocity for this small but finite portion of the 
motion. The data are insufficient to determine whether the 
velocity is variable or constant.) 

Examj^le 2. The same wheel, besides describing 1° in 0.010 
sec, is found to describe 1° 2' in the next 0.01 second. Com- 
pute the angular acceleration, as near as may be from these data, 
for this part of the motion. As before, da = 0.01745 radian, 
and we now have the additional fact that da =z 0.01803 radian, 
each of the time-intervals being dt = 0.010 sec If we substi- 
tute directly in eq. (YH), there results 

0.000581 radian 

= — T— — - = 5.81 rad. per sec. per sec 

(0.01 sec.) 

That is to say, at this part of the wheel's motion its angular 
velocity is increasing at the rate of 5. 81 velocity -units per sec- 
ond, i.e., 5.81 radians per second per second. 

Another method is this: The velocity at the mid-point of 
the da is, as before, 0.01745 -f- 0.01, = 1.74 radians per sec- 
ond, while at the middle of the da' it is 0.01803 -^ 0.01,= 
1.803 radians per sec. ; taking the difference of which we find 
that 0.058 rad. per sec. of angular velocity has been gained 
while the wheel was passing between these two mid-points. 
Hence, dividing this gain by the time of passage, 0.01 sec, we 
obtain 6^ = 0.058 -r- 0.01 = 5.81 rad. per sec per sec. 



APPENDIX. 



NOTES 



ON THE 



Graphical Statics of Mechanism. 



Peepared by I. p. OHUECH, Coenell Univeesity. 

These notes are based mainly on the work of Prof. Herrmann of Aix-la- 
Chapelle, the earlier form of which was presented as an appendix to Vol. 
Ill, of Weisbach's Mechanics. 

It is thought that the form of presentation adopted in the following pages 
is that best adapted for students already familiar with the Graphical Statics of 
quiescent structures, which Prof. Herrmann assumes is not the case with the 
readers of his book. 

For greater clearness, all force polygons have been placed on separate parts 
of the paper from the mechanism itself, instead of being superposed on the 
latter as in Prof. Herrmann's book. 

The figures referred to in the text will be found in the back of this pam- 
phlet; while the paragraphs (§) referred to (of ahighernumber than **30") will 
be found in the writer's *' Mechanics of Engineering." 

For table of contents see p. 28. 



1. Assumptions. The forces acting on each part of any mech- 
anism here treated will be considered to be in the same plane 
and to form a halanced system, i.e., to be in equilibrium ; in 
other words, the motions of the pieces take place without sen- 
sible acceleration (or the effect of inertia is disregarded). (See 
p. 440 of Prof. Un win's Machine Design for consideration of 
the inertia of a piston.) Also, the weights of the pieces will be 
neglected unless specially mentioned. 

=> 2. Graphical Treatment. A " two-force " piece, or '' two-force " 
member of a mechanism is one acted on by only two forces ; 
which forces, therefore, must be equal and opposite and iiave a 



2 NOTES OK THE GRAPHICAL STATICS OF MECHANISM. 

common line of action, for equilibrium. No force polygon 
need be drawn for such a piece. 

A Three-Force Piece. Here, for equilibrium, the three lines of 
action must meet in a common point and the force polygon is a 
triangle (see § 325); for example, see Fig, A (bell-crank), (p. 28). 

For the equilibrium of 2. four-force apiece (Fig. B), the result- 
ant of any two of the forces must be equal and opposite to that 
of the other two forces. The common line of action of these 
resultants is the line joining the intersections a and 5, while their 
common amount is given by the ray OC^ which is parallel to 
a. .h and is a diagonal of the closed force polygon (here a quad- 
rilateral). If the four forces act in parallel lines, and two are 
given, we determine the other two by an "equilibrium poly- 
gon," etc., by § 329 ; i.e., we treat the two unknown forces as 
pier-reactions, even if their action-lines (one or both) are between 
the action-Unes of the known forces. (For example, see [A] in 
Fig. 17, Plate Y, where from the known forces S^ and 8^ we 
construct the two unknown, P and R^ in given action -lines, to 
balance them. The equilibrium polygon begins at c in the 
left-hand abutment-vertical and consists of segments c...d^ 
d. ,.ej and e . . .f terminating in the right-hand abutment- verti- 
cal, aty. We then draw c . . .y as the abutment-line, or " closing 
line.") 

No more than four forces will act on any piece. 

3. Efficiency. In each of the following problems there will be 
but one working force or driving force, P, and but one useful 
resistance, Q ; all other resistances being due to friction. For 
present purposes we are to understand by efficiency the ratio of 
the value P^^ which would be sufficient for the driving force if 
there were no friction of any kind, in order to overcome Q 
(without acceleration), to the value P, which it must actually 
have, to overcome Q under actual conditions, i.e., with friction. 

P 
Hence efficiency — r) — -y^. [Strictly, the efficiency involves 

the distances traversed ; see § 5.] 

4. Backward Motion. If the useful resistance ^ is a load due to 
gravity, i.e., a weight, the value to which the working force 



3 KOTES OK THE GRAPHICAL STATICS OF MECHANISM. 

(now such no longer) must be diminished in order to allow the 
mechanism to run backward without acceleration will be called 
P' ; and if in any case P' is found to be negative^ we recognize 
the mechanism to be self-locJcing ; that is, it will not run back- 
ward (or " overhaul ") when the working force is zero, but, on 
the contrary, a force must be applied in the action-line of the 
working force in the opposite direction to cause a backward 
motion. For instance, in Fig. C,* the force i^ is necessary for 
the uniform downward motion of the handle A, to raise Q, and 
overcome friction at all points. With no friction, P^ would be 

P^ 
sufficient to raise Q, and the efficiency = -W*; whereas, to enable 

Q to sink uniformly, a still smaller force, viz. P\ must be 
applied at A (but still positive in this case, so that the machine 
is not self-locking). 

5. Efficiency in General. . . , with one working force P and one 
useful resistance Q, is the ratio of the work Qs' to the work Ps, 
where s' and s are the respective distances worked through in 
forward motion by Q and P (simultaneously); i.e., 

7;, or efficiency, = -^ (1) 

Now by § 142, if ^{Fs") denotes the sum of the amounts of 
work lost in friction at various points where rubbing occurs in 
the mechanism, we have 

Ps = Qs' + :E{Fs") {a) 

But in the ideal case of no friction (or perfect efficiency), 
taking the same range of motion as before, and letting P„ denote 
the new (and smaller) value of the working force which is now 
sufficient to overcome Q, we have 

P^s=Qs' + {h) 

Hence from {a\ (5), and (1) it is plain that the efficiency may 
be written thus : 

"^-'Ps-Ps-P' • • • • • '^""^ 
the form proposed in § 3 above. 

* On p. 28 of this appendix. 



4 K"OTES OK THE GRAPHICAL STATICS OF MECHAKISM. 

6. Condition of being Self-Locking. If the efficiency is less than 
0.50 and the lost work of friction be assumed to be the same in 
forward as in backward motion (in same range of motion, of 
course), the mechanism is self-locking. 

Proof. For forward motion (no acceleration), letting 2[Fs") 
denote the sum of the amounts of work lost in friction at the 
various points of rubbing, we have, for a definite range of 
motion, (see § 142,) 

Ps = Qs' + 2{Fs"), (2) 

and for backward motion (same range), similarly, Q being a 
a working force and P' a resistance, (see § 142,) 

Qs' = Ps + :S{Fs") , (3) 

As implied in the notation, assume that the friction-work is 
the same in backward as in forward motion of the mechanism. 
From (1) and (2) we deduce 

^' = Qs' + 2{Fs") ; or, 2{F8") ^ Qs' g - l]. . . (4) 

Substituting from (4) in (3) we obtain, finally, 

P' = f [2.00-1] (5) 

From (5) it is evident that, when the efficiency is less than 
0.50, P' is negative; that is, the mQohdim^xn is self -locking. The 
assumption made above as to equality of lost work in forward 
and backward motion is not exactly true for any machine, per- 
haps. In most mechanisms the friction work is somewhat less in 
backward motion, and the proposition is then true if for 0.50 we 
put a smaller value for rj. 

Machines of peculiar design may be constructed with rf greater 
than 0.50, which nevertheless are self-locking ; but in these we 
find that the lost work is greater in backward than in forward 
motion. 

7. Sliding Friction. By § 156 we know that if two rough sur- 
faces slide on each other the mutual action or force between 
them is not normal to the plane of contact, but inclined to the 



5 NOTES Oif THE GRAPHICAL STATICS OF MECHANISM. 

normal at an angle 0, the " angle of friction," on that side of the 
normal ojpjposed to the direction (of relative motion) of the hody 
under consideration. Thus, Fig. 1, Plate I, if the block A is 
sliding toward the right relatively to B (no matter which one, if 
either, is absolutely at rest) the mutual force between them acts 
in the line o . ..h'^ if toward the left, then it acts in the line 
o' . . , y . The pressure of B upon A is from m toward o (or o')^ 
that of A upon B from m toward h (or V). 

8. Example! Mill Elevator. (Plate I, Fig. 2.) The single 
rigid body Ag consists of a platform and vertical side-piece, 
which rubs at J, the left side of the fixed vertical guide G^ and 
also at <3^, the right side of the same. 

Let Q be the combined weight of the load on the platform 
and the platform itself, and P the required pull or tension to be 
applied vertically at J., to maintain a uniform vertical upward 
motion (forward motion, here). Besides the forces P and Q^ 
the body Ac., considered free, is acted on by the forces P^ and 
i?2 at the rubbing surfaces a and J, acting at the proper incli- 
nation (0) from their respective normals ; note which side. 

Evidently A . ,g is a four-force piece. Four action-lines are 
known, but only one amount, that of Q. To find the amounts 
of P, P^, and j^2, we join d (intersection, or "co-point" of P 
and P^) with c that of P^ and Q. On the right of the 
figure we begin the force-polygon by laying off m . .o parallel 
and equal to Q (by scale). After drawing through m a line 
parallel to P^, and through o a line parallel to c . .d (the known 
action-line of the resultant of Q and P^), by the intersection k 
we determine P^= m . . k, and the resultant, o. .1c, oi Q and P^, 
But this resultant should balance P and P^, and therefore close 
a triangle with them in the force-polygon ; hence parallels to P 
and P^ . . through Jc and o respectively, finally complete the force 
quadrilateral m. .n., and fix the values of P and i?i, which can 
now be scaled ofi. 

In this figure, for clearness, a large value, about 20°, has been 
given to 0, so that/*, — tan 0,= about 0.36 ; and from the force- 
polygon we find that for Q = 110 lbs., P is about 140 lbs. If 
there were 7iO friction P^ and P^ would be horizontal, and a 



lO 



6 KOTES 01^ THE GRAPHICAL STATICS OF MECHANISM. 

force Po=Q = 11^ ^bs. would be sufficient, vertical and upward 
at A, to maintain a uniform upward motion (or to permit a uni- 
form downward motion, once started). Hence the efficiency for 

upward (forward) motion, with friction, isff = -p = rj^r = 0.78 or 

about 78 per cent. 

In uniform hacJcward motion with friction, 12^ and i?, (now 
call them ^/ and i?/ ) will change their action-lines to the other 
sides of their normals ; i.e., they will now act along a, .d^ and 
h . .e\ respectively, at an angle with the normals. Hence the 
line G^ . . d\ in backward motion, takes the place of c . .d^in 
getting the new force-polygon (see on left in Fig. 2); i.e., o\ . ¥ 
is drawn parallel to c'. . d\ and m\ . k' parallel to c\ . b, Q 
being known. Then a vertical through ^^ and a parallel to 
d', . a through o' complete the polygon ; which fixes P' as well 
as i?/ and ^/. For Q = 110 lbs., F' is roughly about 80 lbs. 

Since P' is upward (i.e., not negative, being in the same 
direction as P) the machine is not self-locking. To alter the 
design, however, so that it shall be self-locking, we need only 
place the rubbing point a near enough to b to cause a. .d^ to 
pass through, or below, c^; for then o\ . Jc^ will either coincide 
with o\ . n' or pass below it, thus making P' either (zero) 
or negative. But the pressures P^ and P^ will be enormously 
increased. Also, P for upward motion will be larger, and the 
efficiency smaller than before. 

9. Example II. Wedge. (Plate I, Fig. 3.) Kequired the neces- 
sary horizontal force P, at the head of the wedge AP, to raise the 
load Q and overcome the friction at the three points of rubbing, 
a, c, and e. For the upward motion of the block P (wedge mov- 
ing to the right) the lines of pressure at these points of rubbing 
are a . .b, c . .e, and cd; there being no pressure at P. The 
block Pe is a three-force piece, under the action of the known Q 
and the unknown P^ and P^ (i.e., the P^ which acts from g 
toward d) all three action-lines being given. Hence the force- 
triangle is immediately drawn, viz., m . .n . ,o^ and the amounts 
of P^ and P^ become known by scale. 

The wedge AB \^ also a three- force piece, acted on by the R^ 



7 KOTES ON THE GKAPHICAL STATICS OF MECHANISM. 

(now found) pointing from c toward J, and the unknown R^ and 
the required working-force P. Its force-triangle n ..o .,k is 
then drawn, since we have just found R^ and have only to make 
o . .h horizontal (i.e., parallel to P) and n . .h parallel to 
a . .h^io close the triangle and thus determine P =z o , .lc\ (for- 
ward motion with friction). 

With no friction^ P^ is vertical, P^ horizontal, and i?^ follows 
the normal to the plane A . . B \ and the corresponding force- 
triangles, beginning with the known Q^ are m^ . ,n^. . o^ and 
n^ . .0^. .k^\ and thus P^ is found. 

From the drawing we have, with Q — 102 lbs., P = 98 lbs. 

30 
and P^ = about 30 lbs.; so that the efficiency t; = ^ =: 0. 31. 

The angle (p used is about 20° or/ = 0. 36. 

For hachward motion^ R^ w^ould act parallel to c . . d\ 
R^ parallel to ^ . . c\ and R^ parallel io a . . V\ and the corre- 
sponding force-triangles are tu' . . n' , . o' and n' . . o' . . h\ the result- 
ing value of P' being negative. That is, P' must act from right 
to left^ for backward motion. Hence the mechanism is self- 
locking (for these particular data, in Fig. 3). 

Note, — It can easily be shown that if the angle of the wedge 
(i.e., the angle between its two sides) is less than twice the 
angle of friction (supposed the same for all three rubbing 
contacts) the mechanism is self-locking. (This is best shown 
graphically.) 

10. Example III. The Jack-Screw. (Plate I, Fig. 4.) Eequired 
the value of each force P of a couple, in a horizontal plane, applied 
to the cross-bar a^ 5^, in order to raise the load Q and overcome 
the friction between the surfaces of the square-threaded screw 
and nut. The screw-shaft is vertical. Assume no pressure at 
the edges of the threads. The pressures on the helical surfaces 
may be considered concentrated at two points d^ and d^^ diametri- 
cally opposite, in the middle of the width of the thread. The 
pressure at <i„ viz. R^^ will lie in a vertical plane ~\ io a^. . h^^ and 
make an angle cf) -^ a {ow the left) with the vertical, while R^ on 
the other side makes an equal angle with the vertical (but on the 



8 NOTES ON" THE GRAPHICAL STATICS OF MECHANISM. 

right), a is the angle which the helix of screw surface makes 
with the horizontal. 

Projecting the five forces on a plane ~| to the bar a^ . . 5^, we 
find that Q must balance the vertical components of Ii^ and Ii^, 
which justifies the force-triangle drawn at [B], where, with Q 
given, we easily fix I^^ and Ii^ bj drawing w . . m and k. .m at 
an angle of + ^ with Q, as shown. Therefore the horizontal 
component of I^^isp . , m, that of JR^i^ m . ,jp. 

Projecting all the five forces on a horizontal plane, we note 
that the two unknown i^'s must balance the couple formed bj 
the horizontal components of R^ and R^^ as if these components 
(which we will now call R^ and R^ were applied directly to the 
bar a^.,h^ at points vertically above d^ and d^. Hence (see [ C~\ in 
Fig. 4) we treat the known R^ and R^ (each = m . .])) as par- 
allel forces applied perpendicularly to a straight beam a . .h 
supported at a and J on || smooth surfaces whose normals are "1 
to a. .h and || to R^ and R^. The reactions of these supports 
will be II to J?g and R^ and are the forces R and P required. At 
[6*] in Fig. 4:, d^. . d^— the distance d^. .d^ of [J.], and a. . 5 
— distance a^ . . h^. 

Hence, using the construction of § 329, we make x . .y = R^, 



y . . X = R^, select a pole O at convenience, draw the three rays 
. .X, 0. . ?/, and 0. .x, and a corresponding equilibrium poly- 
gon u . . t . . s . , r, beginning at any point u in the action-line 
u. .h oi the lower P. Join u. .r, the abutment-line, and draw 
a line || to it through the pole to ^x n'\ then x . .n' ^= /^re- 
quired. 

With no friction^ — 0, and R^ and R^ are each equal to 
m^ . .^, instead of m..j^\ and P^ is proportionally smaller than 
P. In this figure P^ is about two thirds of P\ i.e., the efiicien- 
cy is about 0.66. 

For hackward motion^ R^ and R^ must act on the other sides 
of their respective normals ; i.e., in \^R\ we should use a — 
instead of «' -]- 0, and if in that case a were less than 0, the 
point m would fall on the right of^, and P' would be negative; 
i.e., the screw would not run backward (would not " overhaul") 
when there was no force on the cross-bar; (self-locking). 



9 KOTES 01^ THE GRAPHICAL STATICS OF MECHANISM. 

11. Pivot Friction. (Plate II, Fig. 5.) Since the frictions at the 
base of a flat-ended pivot (see § 168) are equivalent to a couple in 
which eacli force is i/7?, or i/Q with present notation, and 
whose arm is -J of the radius r (so that the moment of the couple 
is %fQr\ we may suppose the pressure concentrated at two 
opposite points, in the circumference having a radius = f r, as 
in Fig. 5. The pressures at these points are inclined at an angle 
= <p with their respective normals and in the proper directions, 
as shown in Fig. 5. If uniform motion is to be maintained by a 
horizontal couple of a moment = Pa^ P being unknown and Q 
and given, we may proceed graphically as in the preceding 
problem, making <af == (zero); a here denotes the distance a , J) 
of preceding problem. 

12. Journal Friction. Assuming that the journal is fitted to its 
bearing with sufficient play to permit the line of contact to take 
any position (in the circumference of the bearing) called for by 
the conditions of equilibrium (i.e., there is no initial clamping), 
we know from § 163 that the action-line of the mutual pressure 
between them must be tangent to the friction-circle, when rub- 
bing is taking place. As to which side of the friction-circle is to 
have the action-line drawn tangent to it, that is a matter depend- 
ing on the direction of the motion and must be decided by the 
statement in § 7. The bearing in which the journal turns may 
itself be in motion (the crank end of a connecting-rod, for 
instance) and the direction of relative turning must be considered. 
The following examples will bring out clearly all the relations in 
such cases. Each friction-circle will be nmch exaggerated, for 
clearness, in the figures, and must not be mistaken for a journal. 
The radius of a friction circle is r sin 0, where /• is the radius of 
the journal. (IN^ote carefully the relations in § 163.) 

13. Example IV. Bell-crank. (Plate II, Fig. 6.) The load Q is 
given in amount and direction, being supported by a vertical 
link which is jointed at^ to the bell-crank. The working force, 
P^ has a given direction and is applied through a link jointed at 
^to the bell-crank. Both joints consist of journals in bearings. 
P=z'lio overcome friction at A, B^ and C^ and to raise Q. The 
motions at A and B are such that the action-line of Q must be 



10 KOTES 0>f THE GRAPHICAL STATICS OF MECHAKISM. 

tangent to ^'s friction-circle on its left; and that of P tangent 
to ^'s friction-circle on its under side. 

Drawing, then, these tangents || to the respective known di- 
rections of P and Q, we have A., a and c . . (2 as their action-lines, 
meeting at a, through which (since the bell-crank is a three-force 
piece) the third force P must pass (the reaction at (7), and this 
must be tangent to the friction-circle at G on its %ijp^er side. 
Hence a line through a and tangent to the friction-circle at C on 
upper side, is the action line of 7t, viz., a . .h. The force-trian- 
gle h.,m. .n\^ now easily drawn, Q being the known force and laid 
off first, and P ^^ in..n\^ thus determined. With no friction., i^, 
and Q (directions unchanged) must pass through the centres of 
the pins at A and ^, and intersect at a^. P then passes through 
a^ and the centre of (7, i.e., acts along aj)^. P^ = mn^ in the 
new force-triangle, and the efficiency = m . . ?i„ -^ m . . ti. For 
hackward motion, we make the action-lines tangent to their re- 
spective friction-circles on the other side in each case ; then P' 
= m . . n' . — -' 

14. Example V. The Slider-crank. (Plate II, Fig. 7.) The 
wheel IT and the crank B form a single rigid bodj^ turning on a 
fixed bearing G. The connecting-rod or link, BD^ is pivoted 
about the crank-pin at one end and to the cross-head pin at the 
other. The cross-head block slides in a right line between guides, 
and receives the pull (or push) of the piston-rod in the axial line 
of that rod, i.e., through the centre of the pin at D. The resist- 
ance Q being given, applied in line «.. h to wheel TF", required 
the necessary valueofP for uniform motion m the given position 
of the mechanism. The connecting-rod is a two-force piece, and 
in its present position {B on the right of a vertical through G) 
the pressure at the cross-head pin is tangent to the friction-circle 
there on its upper side ; that at the crank-pin on the lower side 
of the friction -circle there. Hence a line drawn so as to be tan- 
gent to the two circles in the manner stated is the action-line of 
R^, the crank-pin pressure (as also that at the cross-head pin); i.e., 
draw e. .h. 

The three-force piece, WGB, is acted on by the known Q, by 
i?j, and a pressure at the bearing (7, viz., P^^ which is tangent to 



11 l^OTES OJq" THE GRAPHICAL STATICS OF MECHANISM. 

the friction-circle at C on the upper side ; therefore i?^ must act 
in the line h . ./ drawn from the intersection h, of Q and II^ tan- 
gent to friction-circle at on upper side. Hence the force-tri- 
angle for WOB is easily completed by making o. .m\\ and = 
Q, and drawing m . . 71 and 0. .n \\to c . .h and /*. . h, respectively, 
thus determining 7?^ and E^. J^ow the cross- head block, D/i, is 
a three-force piece acted on by It^ (pointing toward the left), now 
known ; by the unknown i^ and by the unknown ^, which is the 
pressure coming from the upper guide, and must act in a line d. .e 
through the intersection of the action-lines of I^ and ^j, viz., e, 
and makes angle = (in direction as shown) with the normal to 
the guide surface. Its force-triangle, then, is rks, k . .r being = 
and opposite to m . . ^, s .,r being drawn || to jP and k. .s \\to 
d. .e. Tbus JP is found, being = rs. 

As for i^Q (i.e., with no yriction), R^ would act through the 
pin centres, thus raising h ; R^ would act through h and the centre 
of C \ while j^3 would be normal to the guide surface. "With new 
force-triangles, on this basis, we find P^ = '^••'^05 whence the ef- 
ficiency, = P^ -h Py is found. 

In any other position of the mechanism a similar method is 
available. 

15. Example VI. Beam-engine with Evans's straight-line motion. 
(Plate II, Fig. 8.) WIC3I is a single rigid body (wheel and 
shaft) turning on a fixed bearing at ^ Mis the crank-pin, 3fP 
the connecting-rod, PO Si link turning in a fixed bearing at P, 
and pivoted at C to the beam AOP, one end of which, P, is 
guided in a horizontal right line by the block i^and straight 
guides. The pin C being mid-way (in a straight line) between P 
and A, and the length CP being made equal to AC, which also 
== CP {between centres; i> and ^ at the same level), A will be 
compelled to move in a vertical straight line, as if it were itself 
guided by a straight edge. The vertical piston-rod is linked to 
the beam at A. Required the necessary steam pressure P on the 
piston-head, for upward motion, the resistance Q being applied 
to the wheel W in the line x. .w, and the parts having the 
position shown in the figure. 

The link PCis sl two-force piece, subjected at this instant to 



12 KOTES OIT THE GRAPHICAL STATICS OF MECHANISM. 

some thrust, R^^ whose action-line must be tangent (below) to the 
friction-circle at (7, and also (above) to that at D. Similarly the 
link or connecting-rod BM\^ a two-force piece, under a tension 
^j, whose action-line is tangent to the friction-circle both at M 
and B. At M thistangency is on the right ; at B it may be either 
on the right or the left according as the link has not yet 
reached, or has passed, a certain critical position which is very 
nearly the position chosen for the figure; in which the tan- 
gency has been drawn on the left SLt B. 

The block i^is a two-force piece, under a thrust 7?„ directed, 
as shown, at an angle with the normal to the guide-surface 
below it. 

Construction. "We first draw the force-polygon for WKM, a 
three-force piece, the forces being the known Q in line w .,x^ the 
unknown B^ in the known line c. .h^ and the unknown reaction 
B^ at the bearing K. 

R^ must act in a line a? . . y through the point a?, and tangent, 
on left, to the friction-circle at K. Hence by laying off ^ . . m = 
and II to Q^ and then drawing m . .n \\ to h. .c and o . .71 \\ to 
x..y, we determine R^ and R^. 

The beam ABCB is a four-force piece under the forces i?^, B^, 
^4, and i?5, their action-lines being all known and^^ now known 
in amount, = m . .n. The direction oi R^ becomes known from 
a consideration of the piston and rod which together form a 
three-force piece, being acted on by P, by the stufiing-box pres- 
sure B^, and by B^ reversed. Since B and B^ intersect at a, and 
since B^ must pass through a and be tangent on the right to the 
friction-circle at A, its action-line is easily drawn. Eesuming the 
body ABOB, we find the intersection, e, of B^ and B^ (N.B. e is 
off the paper) and join e with/*, the intersection of B, and B^. 
Making n^ . . m, = and \\to B^ (now pointing down), and draw- 
ing n^. .r\\to f. . e and m^. .r \\to B^\ also r . .u\\to B^ and 
n^..u \\ to B^: we finally determine B^, B^, and B^. 

Having found B^, the force-triangle for the three-force piece 
A . .« (piston, etc.) is easily drawn (see [Z] in Fig. 8) and B finally 
determined. 

With no friction we find B^ by taking friction-circle centres 



13 NOTES ON THE GRAPHICAL STATICS OF MECHANISM. 

instead of tangencies and making R^ and R^ ~| to their respective 
sliding surfaces. 

16. Example VII. Oscillating Engine. (Plate III, Fig. 9.) 
Here the connecting-rod is dispensed with, the piston acting 
directly on the crank-pin J., while the cylinder EE oscillates on 
trunnions turning in fixed bearings. The crank AB and wheel 
TF constitute a single rigid body turning in a fixed bearing B. 
The resistance Q being applied to W in line d . . c, we wish to 
find the proper steam pressure P on the left of the piston to 
overcome Q and all intervening frictions, for motion (with in- 
sensible acceleration), of the mechanism in the position shown. P 
acts centrally along the axis of the piston-rod. The pressure P^ 
is that of the stuffing-box against the piston-rod, and P^ that of 
the cylinder against the edge of the piston. Kemembering that 
the piston and cylinder always have a common axis as they " tele- 
scope" in and out, we see that there are only two forces external 
to these two pieces when considered together, viz., the pressures 
at the crank-pin and at the trunnion, which two pressures {P^ 
must therefore be equal and opposite, both of them acting in the 
line h . .g tangent to the two friction-circles (on the lower side at 
both A and C). We have thus found the action-line of P^, 

]S'ow consider the three-force piece WBA. The forces are 
the known Q in line d. .e, the unknown P^ in the line h. . g, and 
the reaction or pressure P^ in a line a ..e which we draw through 
e, the intersection of P and Q, and make tangent (on the upper 
side) to the friction-circle at the bearing B. Hence, making m ,.n 
= and II to Q^ and m..T and n . .T\\io e ..Ji and e .,a^ respect- 
ively, we ^-^L the amounts of P^ and P^ by this force-triangle 

The piston is a four-force piece, under the action of P^^ now 
known, acting toward the left in line h. .e\ by the unknown P 
acting, along the centre line of the piston-rod ; and by the two 
unknown reactions, P^ and P^^ inclined at angle to their re- 
spective normals, and acting at known points. Hence join 3/, 
the intersection of P^ and ^3, with a?, that of P and P^\ make 6> , . ^ 
= and II to i?2) draw^ . . 5 || to ^3, and o , .s\\io y . . x, to deter- 



IT 



14 KOTES ON" THE GRAPHICAL STATICS OF MECHAKISM. 

mine s; s ,,t\\to P and o . . ^ || to ^^ are then drawn to fix t, P 
can now be scaled off, as also ^3 and R^. 

With no friction^ R^ would act in the axis of the piston 
through the centres of A and C. In place of e we would have e^ 
(not shown in figure), and ^ would act through e^ and the centre 
of ^. R^ and R^ would be "| to their respective rubbing surfaces. 
"We would then obtain R^=. m , ,t^ and P^ = R^ = t/i . . r^ ; 
whence the efficiency is = P^ -^ P =z m . . r^ -^ s . , t. 

16. Example VII. The Blake Ore-crusher. (Plate III, Fig. 10.) 
^and Z are fixed walls, HP oscillates about a fixed bearing Z^^, 
WPa is a wheel and crank rotating on a journal in a fixed bear- 
ing ZI The connecting-rod a,.c communicates motion to the 
two links or struts, PPJ and OR, forming a toggle-joint and 
causing PP^ to oscillate. 

For the given position of the parts, Q being the resistance 
offered by a piece of ore, h, to any further motion of P toward 
the left, required the value of P, applied in the line W. .h to 
wheel WP to overcome Q and the pressures at the seven sockets 
or bearings. 

The two links PP and OP are evidently two-force pieces, PP 
being subjected to some thrust R^ in line A . . <?, OP to some 
thrust R^ in line c .. P ; these lines being drawn tangent to the 
various friction-circles in the manner shown. 

R^ and Q intersect in A, hence the line /* . . A, drawn through 
h and tangent to the friction-circle at P^, must be the action-line 
of 7tj, the reaction at the bearing P^. That is, Q, R^, and R^, 
are the forces acting on the three-force piece PP^^ ; therefore, 
Q and the three action-lines being known, we easily complete 
the corresponding force-triangle . ,7n . .n and thus determine 
R^ and R^. 

Passing to the three-force piece a. . b we have the action-lines 
of R^ and R^ already intersecting at c, and the amount of R^ = 
o . .n. The third force R^ has an action-line passing through c 
and tangent (on left) to the friction-circle at the crank-pin. 
Hence draw c. ,a accordingly, and complete the force-triangle by 
making h . , i \\ and = to 6> . . n, (i.e., to R^, and i .,r and 



15 IS^OTES ON THE GRAPHICAL STATICS OF MECHANISM. 

Ic, .T II to a. .0 and B . , c respectively ; which determines B^ 
and R^. 

Finally, the three-force piece a . . F.,W{^ seen to be acted on 
by ^4, now known ; by the unknown P in line P . ,h\ and by 
the bearing-reaction P acting through h and tangent (on right) to 
the friction-circle at P. Draw h , . F^ then, as indicated, and the 
force-triangle is readily formed, t. . s . ,u^m 2lY1 obvious manner, 
whence s . .^c^ = P^ is scaled. 

Without friction we would as usual draw the action-lines of 
the divers ^'s through the centres of the various journals, or 
sockets, instead of tangent to their friction-circles. With a new 
set of force-triangles on this basis (see broken lines on the right 
in Fig. 10), we obtain P^^ considerably smaller than^P. In this 
exaggerated figure the efficiency is only about 0.50 ; but in Prof. 
Hermann's drawing, with/'^ for journal friction = 0.10, he ob- 
tains a value of 0.80 for the efficiency. 

17. Rolling Friction (so called). This kind of resistance is due 
to the fact that the point of apphcation of the force acting be- 
tween a wheel and the rail or surface on which it rolls is not at 
the foot of the perpendicular dropped from the centre of the 
wheel upon the rail, but a little in front (in direction of rolling) 
by an amount, or distance, h — about 0.02 in. for iron wheels on 
an iron rail, and about 0.50 in. for a wagon-wheel on a dry 
macadamized road (J) — from 2.^^00 to 3. ''00 on soft ground). 
See § 1Y2 ; (also Prof. Eeynolds's article in the Philos. Transac, 
vol. 166.) As to the direction of the pressure of rail on 
wheel it is somewhere within the cone of friction so long as 
perfect rolling (i.e., no slipping) proceeds, being the equal 
and opposite of the resultant of all the other forces acting on the 
wheel. 

Thus, Fig. 11, Plate III, if Q is the weight of the roller, 
applied in the centre o\ the necessary force P^ horizontal and 
acting through the centre of the roller (to maintain a uniform 
rolling motion), is determined by the fact that the resultant of P 
and Q must act through c, and therefore along the line o'. . <?, 
where c . .d — the distance h just mentioned. Hence, making 
the line Q through o equal by scale in length and || to ^ at o\ 



16 NOTES ON THE GKAPHICAL STATICS OF MECHANISM. 

and drawing o . . m, \\ to o\ . c, as well as a horizontal through the 
lower extremity of Q, we determine both I^ and the rail pressure. 

Again, Fig. 12, if the rolling action occurs on both sides of 
the roller, as at c and e when a weighted plank is moved horizon- 
tally (to left, here) on loose rollers, we note that it is a two-force 
piece (neglecting the weight of the roller), the action-line of the 
compressive forces being e . , c, both e and c having been located 
at distance = h from the perpendicular through the centre (in 
the proper directions ; and then, we may have different Z>'s at 
the two points of rolling contact). Here the upper plank is 
moving and the horizontal force (toward left) which must be 
applied to it to maintain this motion uniformly must be equal to 
the sum of the horizontal components of the various inclined ^'s, 
one from each roller, the sum of the vertical components of the 
latter being equal to the total load on the plank. 

In the uniform motion of a car-wheel, J^A, Fig. 13 (brakes 
not on), the (double) wheel and its axle constitute a single, rigid, 
two-force piece, acted on by the rail pressure, or reaction, at c 
(c . .d being made — h) and by the pressure of the bearing against 
the journal or axle at E ; and this pressure must be tangent to 
the friction-circle at E {ow the right for motion here shown, car 
moving to the left). The horizontal component of the equal and 
opposite of this i? is the tractive resistance, on a straight level 
track with uniform motion (besides the resistance of the air), and 
is continually overcome by the tension in the draw-bar of the 
locomotive. R is practically equal to its own vertical component 
which equals the portion G of the car's weight borne by this 
wheel, and this horizontal component of i? is = G tan ^, where 
d denotes the complement of the angle e . .c. . d. 

If the brake is in action, Fig. 7>,^ exerting a pressure P' at the 
point «, this pressure must act along the line <2 . . c* at an angle = 
(p' with the radius a^(note the direction of motion, etc.), pro- 
vided the wheel is not held fast by the brake. Suppose now 
that the wheel, while continuing to roll without slipping on the 
rail is 07i the j^ohit of slipping upon it^ that is, suppose that " skid- 
ding " is impending ; then P'\ the pressure of the rail on the 

______ 



17 NOTES OK THE GRAPHICAL STATICS OF MECHANISM. 

wheel, besides being applied at <?, must take the direction c . . <?, 
at an angle 4>" with the vertical (track level), on the right, here. 
Since P' and P" intersect at o, the only remaining force acting 
on the rigid body ^^TF (which now is a three-force piece) must 
pass through o. This force is P"\ the pressure of the bearing 
on the journal, and it must also be tangent to the friction-circle 
at E. Its action-line, therefore, is easily drawn and is e . . 6>. A 
corresponding force-triangle being now constructed (not shown in 
the figure) with sides || to c . ,o^ a, ,o, and e . .o^ respectively, in 
which the vertical projection of P'^ is made = G, the portion of 
weight of car coming on this wheel (double), we determine the 
value of each force. Since (p" for impending slip (friction of 
rest) is greater than cp' for actual slipping, greater resistance is 
offered by impending than by actual " skidding." In most cases 
in practice the distance c . . fZ, = J, is so small that its effect in 
graphical work is almost inappreciable. 

18. Example VIII. Friction-rollers of Crane. (Plate III, Fig. 14.) 
At T. .X . . Y WQ have a vertical projection of the crane. The 
horizontal pressure of the crane at the base is exerted through 
friction-rollers (shown at P and B) against the side of the fixed, 
conical, and vertical mast e . .h. The vertical pressure induced 
by the suspended load being supported at T^ while a horizontal 
pressure simply ( = ^) is produced at the base X . , F, we shall 
consider the equilibrium of the part G . .E^ carrying the friction- 
rollers, as if it were acted on by the force Q. applied centrally 
and symmetrically in the line a.. Q,.h\ by the pressures R^ 
and E^ of the friction-roller journals against their bearings ; and 
by a force P, applied on the lug <2, and || to face GE. This 
force P is of such an amount, to he determined, as to maintain a 
uniform motion in direction of the dotted arrow. 

That is, EEGI1\& to be treated as a four-force piece. ]N"ow 
each friction -roller is a two-force piece like the car-wheel in Fig. 
13 and the action-line of the compressive forces in each is drawn 
in the same manner as in Fig. 13, noting well the direction of 
motion. We thus determine the action-lines, /"..(^ and d..e^ 
meetmg at o, of the unknown R^ and R^. 

Note. — Of course the force R^ with which the bearing of our 



18 NOTES OX THE GRAPHICAL STATICS OF MECHAi^ISM. 

four-force piece acts against the journal of the friction-wheel is 
the equal and opposite of the R^ with which that journal acts 
against the bearing. The bearing is part of our four-force piece, 
and the force R^ acting upon it is not shown in the figure, but 
has the same action-line as the R^ shown. 

Having now four action-lines and the amount of one force, 
viz., Q^ of the four forces acting on the piece EFGH, we proceed 
to construct the amounts of the other three, in the way so often 
employed, classifying the set of forces into two pairs. Pair off 
P and Q ; they meet at a ; join a ..o ; lay off o". . m — and |i 
to Q ; draw m . . a" \\ to P and o" . .a" \\io o .. a thus fixing 
P, = 7)1. . a." Also draw o" . .n\\io d , . e and a" . .7i\\ to 
o. .y, thus determining the amounts of R^ and R^. 

Since, with no friction, P^ = 0, the efficiency = 0, i.e., all 
the work done by P is wasted (spent in overcoming friction). 

19. Chain Friction. (Plate IV, Fig. 15.) Supposing that the 
groove in the periphery of the pulley A is properly constructed for 
receiving a chain in such a way that, as the chain winds upon the 
pulley, the alternate links place themselves with their planes || 
to or "1 to the axis of the pulley ; then, as each link, c, approaches 
the point ^ where it is to assume its proper place on the circum- 
ference, its neighbor just above being already in place on the 
pulley, and turning with it (whereas the first one remains vertical 
for a time) a turning of one in the hollow of the other^ with 
corresponding friction, is brought about, and in such a direction 
that the pressure between the two is tangent to the friction-circle 
on that side of the latter which {^further from the centre of the 
pulley ; for one turns in the other like a journal in the bearing. 
(The pulley in the figure is turning clock-wise.) 

Similarly, on the other side, at Z, where the chain is unwind- 
ing and thus being straightened, each link turns in the hollow of 
its neighbor, on passing off the pulley, and hence the pressure at 
L is tangent to the friction-circle there on the side nearer to the 
pulley-axle. Again, at the bearing of the journal of the pulley- 
axle, the reaction, R^ is tangent to the friction-circle and on its 
right. 

Let r' denote the radius of the hollow at the extremity of a 



19 KOTES ON THE GRAPHICAL STATICS OP MECHANISM. 

link (equal to the half-thickness of the chain-wire) and r" the 
radius of the journal of the pullej-axle ; also 0' and cp" the 
respective friction-angles at those points. Then the radius of 
the friction-circle at L and 2X K ^=^ a ^=^ r' sin 0', and that at the 
central bearing = ^ =: 7'" sin <p" . Let r denote the distance 
between the centre at K (or L) and that at the middle bearing. 

"We thus see that a weight Q supported on the ascending side 
of the chain (on the left, here) has a lever-arm ^=^ r -\- {a -\-})) 
about the point of application of R / while the vertical force P, 
applied to the descending (and unwinding) side of the chain, has 
a lever-arm of only r — ifb -\- h). Hence for uniform motion we 
have (from equality of movements) 

P : Q \: T + {a-\-l)):r — {a-\-l)). 

From this proportion P may be computed ; or, graphically, 
given § as a load applied to a horizontal beam with supports 
at R and P, we may construct R and P as if reactions of those 
supports, by § 329. Thus, at [B'] in Fig. 15 make m. .7? || and 
= Q ; take any pole O and any point e in the vertical action-line 
of R. Draw the rays . . m, and .. R and the segments 
e.,c^ and c..n\\to them (respectively); join e . .n and make 
..n' Wioit through 0. Then R.,n' = *P, and n'. .m ^ R. 

With no friction P^ = Q, and we can find the efficiency. 

For backward motion the lines of action of the three forces 
have their friction-circle tangencies each on the side opposite to 
that in forward motion, and we have 

P' : Q :: T - {a + 1) : r + {a -\- I). 

20. Example IX. Pulley Blocks or Tackle. (Plate lY, Fig. 16.) 
Let C be the upper block of a tackle of two blocks with three 
pulleys in each (the pulleys turning independently on a common 
bolt). 6^ is suspended from a support and is stationary, and R 
is the tension in the rod or shank by which it hangs. 

The lower block is movable and carries a load Q^ whose uni- 
form upward motion is to be maintained by the application of a 
proper force P or S^, to the chain at the extremity h. The chain 
passes continuously over the six pulleys or sheaves in the manner 
shown in the figure, the other extremity being attached to the 



20 i^OTES OK THE GRAPHICAL STATICS OF MECHA:N^ISM. 

support above. Each straight part of the chain will be con- 
sidered vertical, and all these straight parts have different tensions. 
Denote these tensions by S^ {= P) S^, S^, xS^, S^, S^, and S^. 

For simplicity (as will be seen) let us suppose JP given and Q 
required. Since each pulley of this figure is similarly circum- 
stanced to the one in Fig. 15 where the tension on the unwinding 

• 1 IT 11.1 . r-\-(a -\-h ) 
side was sjreater than that on the other m the ratio y- — r-ri 

(where a, r, and h have the same meanings as in the preceding 
paragraph) we see that 

S,:S/.:r-{a + h) :r + (a + hy, 
S,: S,::r-{a + h): ^ + (a+J); 
S,: S,::r-{a + h):r + la+hy, 
and so on, up to 

S,:S,::r-{a + i):r + {a+l). 

Hence we adopt the following construction : layoff on a hori- 
zontal line, o , .w =^r — {a-{-h); o . .w' = r -{-{a -\- h) ; and also 
w' . .o' =^ r — (a +5). Draw a vertical through each of the 
points c. w^ o' ^ and w' . On the vertical through w^ lay off to 
scale w . .171 = P = S^, and draw and prolong m . . o' to intersect 
the vertical through w' in some point m^; whence w' m' = S^, 
Then a line joining m! with <9, by its intersection with the verti- 
cal through w, gives a point n such that w . .n = /S^; while 
n. .0^ prolonged cuts the vertical through w^ in some point n\ 
giving w'n^ = 8^ ; and so on, until all the six tensions S^ to aS!, 
are found. 

Proof. (By similar triangles we see that the proportions men- 
tioned above are satisfied.) 

We then have (see [6^] in figure), 

Q=^ S, + S,+ S,+ S, + S,+ S,; as also 
P=^S, + S, + S, + S, + S, + S,. 

p 

Having thus found the ratio of P to Q^ i.e., the value of -^ , 

we can easily compute i^ if ^ is given, as in a practical case ; and 
also the efficiency, since with no friction P^ = \Q (for the 
tackle in this case). 



21 KOTES OK THE GEAPHICAL STATICS OF MECHAI^ISM. 

In hackward m,otion^ having P =z S^ given, we would draw a 
line from "tn through o, instead of through o', to fix on', and re- 
turning, a line through m' and o' to fix n^ and so on, each tension 
so found being greater than the one preceding it. And finally we 
would have Q == '^'g -|- ^3 + ^^4 H~ ^5 H~ '^e "f" St> ^^^ the ratio of 
the given Q to the unknown P' thus fixed ; and P' would be 

found fron. the relation : ^= s.+ s, + sX S^+S, + S ; 

21. Example X. The DiiFerential Pulley. (Plate Y, Fig. 17.) 
This is a tackle in which the upper block carries only one pulley, 
which, however, has two grooves in |I planes, but with slightly 
different radii. Also, since friction in the grooves is not sufficient 
for the purpose, projecting pegs or ridges (or some similar device) 
are provided to prevent the chain from slipping in either groove. 
The lower block P carries an ordinary pulley of one groove, the 
weight Q being suspended from the axle of this lower block. 

The chain is endless^ passing twice over the pulley A (once 
in each groove) and once under the lower pulley, while a portion 
hangs freely, as shown. 

Given the load Q^ required the vertical force (or load perhaps) 
P, to be applied to the chain where it unwinds from the otttside 
groove (see figure) in order to raise Q and overcome all friction. 
The lower pulley is then acted on by the vertical force Q^ and 
the two vertical and upward tensions 8^ and S^ each of these three 
forces being tangent to its own friction -circle, as shown, on the 
proper side (note that S^ is on the unwinding side). We find S^ 
and S^ by treating ^ as a load (§ 329) resting on a horizontal 
beam supported in verticals a and n. At [B'] is the force-diagram, 
while a . .h . .n\^ the equihbrium polygon. (See § 329.) 

S^ and -5*2 having thus been determined, we consider the upper 
pulley, which is acted on by four || forces, viz., the known S^ and 
S^, the unknown Pand the unknown R or reaction at the journal 
of the pulley. The four action-lines are known, being vertical 
and tangent to the respective friction-circles m the manner shown. 
Note the direction of the uniform motion (to raise the load Q). 

Again we employ § 329, regarding J. as a horizontal heara or 
lever with supports in the verticals/* and c, and loaded with S^ 



12 



22 NOTES 02ii THE GRAPHICAL STATICS OF MECHANISM. 



and xS'a, both known. We hj oS r . . s = S^, and s . .t=^S^, and 
take any pole 0, drawing rajs from to r, s, and t. From any 
point c in the action-line of 12 (the left-hand reaction, or support- 
ing force) we draw a line || to 6^ . . r to find d, then 6? . .e\\to . .s 
to find e, and a line parallel to O . .t through e, to find / in the 
action -line of the riglit-hand supporting force, P. Drawing c . .y, 
a line parallel to it through fixes n' on the load-line (produced), 
giving in' = P, and n'w = R ; i.e., \A\ is the force-diagram. 

Without friction^ the vertical action-lines would be drawn 
through the centres of the friction-circles, and anew construction 
on this basis would give P^, whence the efl&ciency P^ -^ P can 
be found. 

For haohward motion each force-vertical shifts over to the 
opposite side of the friction-circle from that shown in Fig. 17, 
and the result of a third construction is P'. If P' is found 
to be 7iegatwe^ that is, if n' occurs above t in diagram A^ the 
mechanism is self-loching^ as should be the case in the practical 
machine itself. 

22. Rigidity of HempEopes. Here, as with chains, the effect of 
the rigidity is to cause the tension where the rope is winding on 
to have a lever-arm about the centre of the pulley ^^r-\-a^ where 
r =z radius of circle formed by the axis of the rope when wound 
on the pulley, and <^ = a small distance which from Eytelwein's 
formula for rigidity of hemp ropes may be put = O.OO^Zd^^ where 
d is the diameter of the rope in ynillhnetres^ and whence a will 
be obtained in millimetres. 

The tension on the unwinding side has a lever-arm of ^ — a. 
Hence, having computed a, we deal with hemp ropes as with 
chains. The phenomena observed with wire ropes are different. 
(See § 176.) 

23. Tooth Friction in Spur Gearing. (Plate Y, Fig. 18.) 
This figure shows one gear-wheel driving another, both provided 
with " involute teeth " by which we are to understand that the 
normal a^ . . o^ or o . . a^^ at the point of contact always passes 
through 6>, the intersection of the line of centres with the pitch- 
circle, as motion proceeds. 

We assume here that two pairs of teeth are always in contact. 



23 KOTES ON" THE GRAPHICAL STATICS OF MECHANISM. 

Just now these points of contact are at a^ and a^, and have there- 
fore a common normal a^oa^. 

Rubbing occurs both at a^ and a^^ and evidently in such direc- 
tions that the pressure at a has a^ h^ as action line ; and that at a^ 
has h^a^ as action -line, making the angle of friction with the 
respective normals (or common normal, rather). This common 
normal a/)a^ will be assumed as making an angle of 75° with the 
line of centres at all times (property of the kind of teeth used). 
Hence the resultant action of the two driving teeth upon those 
driven is represented by an ideal force i?, the resultant of the 
pressures at a^ and a^, and acting through o\ making an angle of 
75° with the line of centres. Notice the position of this angle 
with reference to the direction of motion and to the driven wheel ; 
also that the effect of friction is to cause the action-line of B^ 
which without friction would act dXong a^o a^, to be shifted |I to 
itself a distance (9C>^ farther from the centre of the drivi?ig wheel. 
This distance oo\ can easily be determined by drawing the parts 
concerned on a convenient scale, and will be called C in the next 
paragraph. 

24. Example XI. Pinion Spur-Wheel, Drum and Weight. 
(Plate Y, Fig. 19.) The weight Q hangs by a chain or rope from 
the drum ^ which forms a rigid body with the spur-wheel H, 
with which the pinion A gears. A drives H^ and it is required to 
find what force P, applied to a crank d (forming one piece with 
the pinion) and acting (at this instant) in the line J . . d, will main- 
tain uniform motion ; i.e., overcome all frictions and raise Q 
without acceleration. 

Since A drives ^with tooth-gearing (involute and of same 
design as in preceding paragraph), the line of action of the result- 
ant pressure JR^ between them is ho^g, making an angle of 75° 
with the line of centres, as shown (note on which side), and is 
draw^n through the point o' on the line of centres but at a distance 
= C farther from the centre of the driving pinion A than a point 
in the pitch-circle of the latter. Call this force H^. 

The reaction at the bearing s is some force i?, whose action- 
line must pass through h, the intersection of the action-lines of 
the other two forces, 7?^ and P (since J. is a three-force piece), 



24 NOTES ON THE GKAPHICAL STATICS OF MECHANISM. 

and be tangent (on the right) to the friction-circle at s. The ac- 
tion-line of Q is vertical, and is tangent (on the left) to a friction- 
circle Sit a { just as in Fig. 15 a similar relation holds at IT) ; it 
cuts g . .h £it g, and therefore a line drawn through g, and tangent 
(on right) to the friction circle at IC^ is the action-line of B^, the 
reaction at the bearing IC, 

We thus have the action-lines of all three forces acting on 
each of the three-force pieces, A and J7, while the force Q is 
given. Hence, the force-triangle r.,Tn..n is easily drawn for 
piece H, and determines I^^ and I^^. With n^'m^^ = and |j to 
m. . :^^ as a known side, we then complete the force-triangle for 
piece Ay from which m" . ,t" = i^ is scaled off. 

Without friction, R^ would shift to the position g^ , . h^, E^ 
would pass through h^ and the centre of the circle at s. R^ would 
pass through the centre of the circle at R -and the point g^, in the 
new vertical action-line of Q (through centre of friction-circle 
mentioned above), ^o • • ^o is parallel to g ,,h and passes through 
the intersection o of the line of centres c, ,s, and the pitch-circle 
of the pinion A. Drawing the dotted force-triangles on this basis, 
Q being given, we finally obtain P^ =■ rrh'\, . rj'. The efficiency 
can now be obtained, =^P^ -r- P. 

25. Belt Gearing. In Fig. 20, Plate YI, we have a pulley 
turning in a fixed bearing and driven by a force P, By belt 
connection this pulley drives another, not shown in the figure. 
The tension S^^ on the driving side is greater than that, S^, on the 
following side. If Z is the (ideal) resultant of S^ and S^, then 
the reaction R of the bearing must act in a line through h, the 
intersection of P and Z and tangent (above) to the friction-circle 
at the bearing ; i.e., it acts along a. .h.^ If we assume that the 
helt is on the point of slipping on the smaller of the two pulleys, 
we have the relation 

8n = S.ef, (§170) 

wherey= coefiicient of friction, e is the Naperian Base, and a = 
arc of contact on the smaller pulley in 7t measure, or in radians. 
Although Sn and /S^ are both unknown at the outset, we have 
^ By mistake, B has been drawn in Fig. 20 along a' . .h, instead oi a . .b. 



2o NOTES ON THE GRAPHICAL STATICS OF MECHANISM. 

tlieir ratio from the above equation, and hence can construct the 
action-line of Z(for impending slip only, it must be remembered), 
their resultant, thus determining the point h in Fig. 20 and 
ultimately H and Z, as will be seen. With Z found, we can 
obtain S^ and S^. The value of this ratio, e^"-^ having been 
computed for a range of values of/" and of a, the results may be 
embodied graphically in the spirals shown in Fig. 21, Plate YI, 
these being drawn in such a way, all starting from the point A 
in the circumference of the circle A . , C, that if OA^ the radius, 
represent the smaller tension, S^, and the special value a — AOQ 
in any case be laid off and the radius OC produced till it inter- 
sects the spiral corresponding to the coefficient f proper to the 
case in hand, B being this intersection ; then OB = S^, and 
OB = Sn —S^ (which multiplied by the velocity of the belt 
gives the power transmitted). Or, whatever S^ may be, the 
ratio BO : AO =^ the 7'atio S^ : S^, and may be obtained from 

the diagram if y and a are given. [N.B. — Note carefully that 

o 
the relation -^ = 6-^" only holds when the belt is actually 

slipping on the pulley -rim (and then / is the coefficient of fric- 
tion of motion) or is 07i the "point of slipping (and theny = co- 
efficient of friction of rest), and is never to he used except for 
those conditions. Of course, in most machinery impending slip 
IS to be avoided, and the only use of the above formula in such 
cases, is to find the ideal maximum value, 6-^'^, for the ratio 
Sn : /iS'„, which the actual value should not approach if slipping is 
not to occur. For the uniform motion of an "idle pulley,"' 
Ignoring axle friction, Sn : S^ is always equal to 1.00. 

26. Example XII. Brake Strap and Drum. (Plate VI, Fig. 22.) 
A is a lever with a fixed fulcrum or bearing at B and has 
attached to it both ends of the belt or strap which passes over a 
pulley and serves as a brake to prevent the acceleration of the 
descending weight Q. The chain sustaining Q unwinds from 
the drum C, rigidly attached to the pulley, which turns on a 
fixed bearing B. Required the proper force P, in a given 
action-line r , . P, to be applied to the lever A, to preserve a 
uniform motion for Q (downward). 



26 NOTES ON THE GRAPHICAL STATICS OF MECHANISM. 

Evidently, from the direction of motion, the tension in the 
strap at d is the greater, = Sn^ and that in the portion a? . . 6 is 
the smaller, = S^. Since in this case there is actual slipping of 
the drum under the strap, the ratio S^ : S^ is known from Fig. 
21, for (saj)/" = 0.18 and «f = |;r (corresponding to 270°), to 
be 2.51. Hence, from the intersection, ^, of the two straight 
portions of the belt, we lay off any convenient distances a . .e 
and a . . d^ such that ad = 2.51 ae, and complete a parallelogram 
upon them as sides ; then the diagonal a . .0 h the action-line of 
Z, the unknown resultant of S^ and S^. 

The pulley and drum, therefore, constitute a three-force piece 
acted on by Q in the vertical line tangent (on the inside) to a 
friction-circle at ^ ; by the (ideal) Z in line a, ,o\ and by the 
reaction, R^ of the bearing B, tangent (on left) to the friction- 
circle and in a line which must pass through g' (intersection of 
the other tw^o force-lines). Knowing Q and all three action-lines 
mentioned, the force-triangle h,.m, .n determines E and Z\ i.e., 
n. .k and m .,n. As for the lever A, though there are really 
four forces acting on it, viz., P^ S^, S^, and R^ (the reaction at Z>), 
we may treat it as a three-force piece, since S^ and S^ are equiva- 
lent to the force Z, now known both in amount and position. 
Hence we draw Ji . .f ■= and || to Z, then f ..g\\io r , . s (since 
Z and P meet at r) and h ,,g \\ to r .. P^ thus fixing P^ = 
/' .. g \\ and P =. g . ,li. Then by resolving Z along a . ,d and 
<2 . . 6 w^e find S^ and S^. {Note. — Q is the working force in this 
example, and P neucral. All work is spent in friction.) 

27. Example XIII. Transmission of Power through two Pulleys 
having Belt Connection. (Plate VI. Fig. 23.) Let the pulley 
A drive the pulley P with uniform motion. At this 
instant the useful resistance is Q acting on B in lin« 7)i . ,q. 
Given Q^ required P^ acting in line h , .a on pulley A^ to over- 
come Q and the journal frictions at B and B'. We assume that 
the belt is on the point of slipping on the circumference of the 
smaller pulley, B] then S^ = S^ 6-^% where a is the arc of con- 
tact on the small pulley. Prolong the action-lines of S^ and /S^ 
to their intersection o. From Fig. 22 find the ratio of S^ to S, 



27 KOTES ON THE GKAPHICAL STATICS OF MECHAKISM. 

for the given coefficient of friction and the arc a, and lay off 
o . . I and <9 . . ^ in this ratio {ok > ol) at convenience. This gives 
6> . . ^ as the action-line of Z, the (ideal) resultant of S^ and xS^^ ; 
of course, that there may be no slip or any approach to it, actual 
values must be secured for these tensions greater than those to 
be found by this construction, which are for impending slip on 
small pulley (and this means the assumption of a less value for 
the ratio S^ : S^. (See p. 186.) 

^ is a three-force piece (so considered here) under the action 
of Z (ideal), P, and Ji^ the bearing reaction. Pulley £ may 
also be treated as a three-force piece under action of Q, of Z 
(reversed) and B, the bearing reaction at £. P and Z intersect 
at J, Q and Z at g. Hence R^ acts through h and tangent (above) 
to its friction-circle ; while R acts through g and is tangent (on 
right) to friction -circle at B. 

Beginning, then, with pulley B^ since the force Q is given, we 
close the triangle "o . . w . . x\xi an obvious manner, obtaining R and 
Z. For pulley A^ now that Z is found, we complete the force- 
triangle y . .3 ,.d^ and determine R^ and P. 

Without friction at the hearings^ R and R^ would pass through 
the centres of their friction-circles and the dotted force-triangles 
would result, whence we have P^ = z..d\ 

To find the belt tensions (for impending slip on smaller pul- 
ley) we resolve the force Z\\ to their directions; see lower part 
of the figure. 

Note. — As far as finding the value of P alone is concerned, 
having Q given, any line whatever could be taken through the 
point o, as the action-line of the resultant of the two tensions, if 
the friction at the bearings were disregarded, and the construc- 
tion would result in the same value of P, whatever the belt- 
tensions, provided the belt did not slip. 

28. Final Remark. From an inspection of the preceding ex- 
amples involving the effect of friction in the working of machines, 
it becomes apparent, as should be expected, of course, that in 
every case this effect is to put the working force at the greatest 
possible disadvantage^ thus exacting as large a value as possible 



NOTES OK THE GRAPHICAL STATICS OF MECHAN"ISM. 5^8 

for it; and from this general principle we maj often decide quickly 
in the matter of tangencies to friction-circles, inclination of a 
pressure on one side or the other from the normal, etc. 




Q Fig. A 






Am. Bk. Note Co. N. T. 



CONTENTS. 



Assumptions 1 

Efficiency 2, 3 

"Overhauling" = ., 3, 4 

Sliding Friction 4 

Mill Elevator 5 

Wedge 6 

Jack-screw 7 

Pivot and Journal Friction 9 

Bell-crank... ^ 9 

Slider-crank 10 

Beam-engine 11 



Oscillating Engine 13 

Ore-crusher 14 

Rolling Friction 15 

Crane (rollers of) 17 

Chain Friction 18 

Tackle 19 

Differential Pulley 21 

Rigidity of Ropes 22 

Spur Gearing 22 

Belt Gearing 24 

Brake-strap and Drum 25 



PLATE I. Figures 1 to 4 . 




Am. Bk. Note Co. N. Y. 



PLATE II. Figures 5 to 8 



k BELL-CRANK 




oVA Fixed 



////'///^//////V/7/////7/////'//////////// 
Sliding 




! {Force Polygons] 



13 



PliATE in. Pigures 9 to 14. 




af^ R 



trunnion -J ^~^g " ^^Im 

OSCILLATING ENGINE. 




Horizontal Section of Stvinging Crane and 
Fixed 3Iast, about which it turns. This section 
is at the base Yin the horizontal plane XT. 
The pivot friction at Tis not considered. 

Am. Bk. Note Co. N. T. 



PI.ATE IV. Figures 15 and 16. 




\ i 

"\A\\\ 



m 



r- (a+.b) 



[B] 






^nV 



\ 
\ 

\ 



\.. 



Am. Bk. Nvtc Co. N. T. 



PLATE V. Figures 17 to 19. 



~X/ Fig, 17, ^<icAv_^ 




PLATE VI, Figures,20 to 23, 




LOGAEITHMS (BEIGGS'). 


N 





1 


2 


3 


4 


5 


6 


7 


8 


9 


Bif. 


lO 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


42 


11 
12 


0414 


0453 


0492 


0531 


0569 


0607 


0645 


0682 


0719 


0755 


38 


0792 


0828 


0864 


0899 


0934 


0969 


1004 


1038 


1072 


1106 


35 


13 


1139 


1173 


1206 


1239 


1271 


1303 


1335 


1367 


1399 


1430 


32 


14 


1461 


1492 


1523 


1553 


1584 


1614 


1644 


1673 


1703 


1732 


30 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


28 


16 


2041 


2068 


2095 


2122 


2148 


2175 


2201 


2227 


2253 


2279 


26 


17 


2304 


2330 


2355 


2380 


2405 


2430 


2455 


2480 


2504 


2529 


25 


18 


2553 


2577 


2601 


2625 


2648 


2672 


2695 


2718 


2742 


2765 


24 


19 


2788 


2810 


2833 


2856 


2878 


2900 


2923 


2945 


2967 


2989 


22 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


21 


21 


3222 


3243 


3263 


3284 


3304 


3324 


3345 


3365 


3385 


3404 


20 


22 


3424 


3444 


3464 


3483 


3502 


3522 


3541 


3560 


3579 


3598 


19 


23 


3617 


3636 


3655 


3674 


3692 


3711 


3729 


3747 


3766 


3784 


19 


24 


3802 


3820 


3838 


3856 


3874 


3892 


3909 


3927 


3945 


3962 


18 


25 


3979 


3997 


4014 


4031 


4048 


4065 


4082 


4099 


4116 


4133 


17 


26 


4150 


4166 


4183 


4200 


4216 


4232 


4249 


4265 


4281 


4298 


16 


27 


4314 


4330 


4346 


4362 


4378 


4393 


4409 


4425 


4440 


4456 


16 


28 


4472 


4487 


4502 


4518 


4533 


4548 


4564 


4579 


4594 


4609 


15 


29 


4624 


4639 


4654 


4669 


4683 


4698 


4713 


4728 


4742 


4757 


15 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


14 


31 


4914 


4928 


4942 


4955 


4969 


4983 


4997 


5011 


5024 


5038 


14 


32 


5051 


5065 


5079 


5092 


5105 


5119 


5132 


5145 


5159 


5172 


13 


33 


5185 


5198 


5211 


5224 


5237 


5250 


5263 


5276 


5289 


5302 


13 


34 


5315 


5328 


5340 


5353 


5366 


5378 


5391 


5403 


5416 


5428 


13 


35 


5441 


5453 


5465 


5478 


5490 


5502 


5514 


5527 


5539 


5551 


12 


36 


5563 


5575 


5587 


5599 


5611 


5623 


5635 


5647 


5658 


5670 


12 


37 


5682. 


5694 


5705 


5717 


5729 


5740 


5752 


5763 


5775 


5786 


12 


38 


5798 


5809 


5821 


5832 


5843 


5855 


5866 


5877 


5888 


5899 


11 


39 


5911 


5922 


5933 


5944 


5955 


5966 


5977 


5988 


5999 


6010 


11 


40 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


11 


41 


6128 


6138 


6149 


6160 


6170 


6180 


6191 


6201 


6212 


6222 


10 


42 


6232 


6243 


6253 


6263 


6274 


6284 


6294 


6304 


6314 


6325 


10 


43 


6335 


6345 


6355 


6365 


6375 


6385 


6395 


6405 


6415 


6425 


10 


44 


6435 


6444 


6454 


6464 


6474 


6484 


6493 


6503 


6513 


6522 


10 


45 


6532 


6542 


6551 


6561 


6571 


6580 


6590 


6599 


6609 


6618 


10 


46 


6628 


6687 


6646 


6656 


6665 


6675 


6684 


6693 


6702 


6712 


9 


47 


6721 


6730 


6739 


6749 


6758 


6767 


6776 


6785 


6794 


6803 


9 


48 


6812 


6821 


6830 


6839 


6848 


6857 


6866 


6875 


6884 


6893 


9 


49 


6902 


6911 


6920 


6928 


6937 


6946 


6955 


6964 


6972 


6981 


9 


50 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


9 


51 


7076 


7084 


7093 


7101 


7110 


7118 


7126 


7135 


7143 


7152 


9 


52 


7160 


7168 


7177 


7185 


7193 


7202 


7210 


7218 


7226 


7235 


8 


53 


7243 


7251 


7259 


7267 


7275 


7284 


7292 


7300 


7308 


7316 


8 


54 


7324 


7332 


7340 


7348 


7356 


7364 


7372 


7380 


7388 


7396 


8 






N 


'. B. — NaperJan log = 


= Brigg 


s'log 


X 2.302. 










Base of Naperian syste 


)m = e 


= 2.71828. 















LOGARITHMS (BEIGGS^. 








N 





1 


2 


3 


4 


5 


6 


7 


8 


9 


Dif. 


55 


7404 


7412 


7419 


7427 


7435 


7443 


7451 


7459 


7466 


7474 


8 


56 


7482 


7490 


7497 


7505 


7513 


7520 


7528 


7536 


7543 


7551 


8 


57 


7559 


7566 


7574- 


7582 


7589 


7597 


7604 


7612 


7619 


7627 


8 


58 


7634 


7642 


7649 


7657 


7664 


7672 


7679 


7686 


7694 


7701 


7 


59 


7709 


7716 


7723 


7731 


7738 


7745 


7752 


7760 


7767 


7774 


7 


(50 


7782 


7789 


7796 


7803 


7810 


7818 


»7825 


7832 


7839 


7846 


7 


61 


7853 


7860 


7868 


7875 


7882 


7889 


7896 


7903 


7910 


7917 


7 


62 


7924 


7931 


7938 


7945 


7952 


7959 


7966 


7973 


7980 


7987 


7 


63 


7993 


8000 


8007 


8014 


8021 


8028 


8085 


8041 


8048 


8055 


7 


64 


8062 


8069 


8075 


8082 


8089 


8096 


8102 


8109 


8116 


8122 


7 


05 


8129 


8136 


8142 


8149 


8156 


8162 


8169 


8176 


8182 


8189 


7 


66 


8195 


8202 


8209 


8215 


8222 


8228 


8235 


8241 


8248 


8254 


7 


67 


8261 


8267 


8274 


8280 


8287 


8293 


8299 


8306 


8312 


8319 


6 


68 


8325 


8331 


8338 


8344 


8351 


8357 


8363 


8370 


8376 


8382 


6 


69 


8388 


8395 


8401 


8407 


8414 


8420 


8426 


8432 


8439 


8445 


6 


70 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


6 


71 


8513 


8519 


8525 


8531 


8537 


8543 


8549 


8555 


8561 


8567 


6 


72 


8573 


8579 


8585 


8591 


8597 


8603 


8609 


8615 


8621 


8627 


6 


73 


8633 


8639 


8645 


8651 


8657 


8663 


8669 


8675 


8681 


8686 


6 


74 


8692 


8698 


8704 


8710 


8716 


8722 


8727 


8733 


8739 


8745 


6 


75 


8751 


8756 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8802 


6 


76 


8808 


8814 


8820 


8825 


8831 


8837 


8842 


8848 


8854 


8859 


6 


77 


8865 


8871 


8876 


8882 


8887 


8893 


8899 


8904 


8910 


8915 


6 


78 


8921 


8927 


8932 


8938 


8843 


8949 


8954 


8960 


8965 


8971 


6 


79 


8976 


8982 


8987 


8993 


8998 


9004 


9009 


9015 


9020 


9025 


5 


80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 


5 


81 


9085 


9090 


9096 


9101 


9106 


9112 


9117 


9122 


9128 


9133 


5 


83 


9138 


9143 


9149 


9154 


9159 


9165 


9170 


9175 


9180 


9186 


5 


83 


9191 


9196 


9201 


9206 


9212 


9217 


9222 


9227 


9232 


9238 


5 


84 


9243 


9248 


9253 


9258 


9263 


9269 


9274 


9279 


9284 


9289 


5 


85 


9294 


9299 


9304 


9309 


9315 


9320 


9325 


9330 


9335 


9340 


5 


86 


9345 


9350 


9355 


9360 


9365 


9370 


9375 


9380 


9385 


9390 


5 


87 


9395 


9400 


9405 


9410 


9415 


9420 


9425 


9430 


9435 


9440 


5 


88 


9445 


9450 


9455 


9460 


9465 


9469 


9474 


9479 


9484 


9489 


5 


89 


9494 


9499 


9504 


9509 


9513 


9518 


9523 


9528 


9533 


9538 


5 


90 


9542 


9547 


9552 


9557 


9562 


9566 


9571 


9576 


9581 


9586 


5 


91 


9590 


9595 


9600 


9605 


9609 


9614 


9619 


9624 


9628 


9633 


5 


92 


9638 


9643 


9647 


9652 


9657 


9661 


9666 


9671 


9675 


9680 


5 


93 


9685 


9689 


9694 


9699 


9703 


9708 


9713 


9717 


9722 


9727 


5 


94 


9731 


9736 


9741 


9745 


9750 


9754 


9759 


9763 


9768 


9773 


5 


95 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 


5 


96 


9823 


9827 


9832 


9836 


9841 


9845 


9850 


9854 


9859 


9863 


4 


97 


9868 


9872 


9877 


9881 


9886 


9890 


9894 


9899 


9903 


9908 


4 


98 


9912 


9917 


9921 


9926 


9930 


9934 


9939 


9943 


9948 


9952 


4 


99 


9956 


9961 


9965 


9969 


9974 


9978 


9983 


9987 


9991 


9996 


4 






N. B.— 


Naperi 


an log : 


= Briggs' log 


X 2.302. 











Base of 


Naperian Sys 


tern = 


e ■= 2.71828. 









TRIGONOMETRIC RATIOS (Natural) 


; including 


<■ '"arc," by 


which is meant the " tt- | 


measure 


" or ' 


'circular measure " of the ang 


e; e.g., arc 100° = i 


.7453293. 


= m 


of 7t. 


arc 


degr. 



sin 


CSC 


tan 


cot 


sec 


cos 


90 




o.ooo 


0.000 


inf. 


0.000 


inf. 


I.OOO 


I.OOO 


I.571 


0.017 


I 


0.017 


57.3 


0.017 


57.3 


1. 000 


I.OOO 


89 


1.553 


0.035 


2 


0.035 


28.7 


0.035 


28.6 


1. 001 


0.999 


88 


1.536 


0.052 


3 


0.052 


I9.I 


0.052 


19.1 


1. 00 1 


0.999 


87 


1.518 


0.070 


4 


0.070 


14-3 


0.070 


14.3 


1.002 


0.998 


86 


1. 501 


0.087 


5 


0.087 


II-5 


0.087 


II. 4 


1.004 


0.996 


85 


1.484 


0.105 


6 


0.105 


9.6 


0.105 


9.5 


1.006 


0.995 


84 


1.466 


0.122 


7 


0.122 


8.2 


0.123 


8.1 


1.008 


0.993 


83 


1.449 


0.139 


8 


0.139 


7.2 


0.I4I 


7.1 


1. 010 


0.990 


82 


1.432 


0.157 


9 


0.156 


6.4 


0.158 


6-3 


1. 012 


0.988 


81 


1. 414 


0.174 


10 


0.174 


5.8 


0.176 


5-7 


1. 015 


0.985 


80 


1.396 


0.192 


II 


O.191 


5-24 


0.194 


5.14 


1. 019 


0.982 


79 


1.379 


0.209 


12 


0.208 


4.81 


0.213 


4.70 


1.022 


0.978 


78 


I.361 


0.227 


13 


0.225 


4.45' 


0.231 


4.33 


1.026 


0.974 


77 


T.344 


0.244 


14 


0.242 


4-13 


0.249 


4.01 


1. 031 


0.970 


76 


1.326 


0.262 


15 


0.259 


3.86 


0.268 


3-73 


1.035 


0.966 


75 


1.309 


0.279 


16 


0.276 


3-^3 


0.287 


3.49 


1.040 


0.961 


74 


1. 291 


0.297 


■ 17 


0.292 


3.42 


0.306 


.3.27 


1.046 


0.956 


73 


1.274 


0.314 


18 


0.309 


3.24 


0.325 


3.08 


1. 051 


0.951 


72 


1.257 


0.332 


19 


0.326 


3.07 


0.344 


2.90 


1.058 


0.946 


71 


1.239 


0.349 


20 


0.342 


2.92 


0.364 


2-75 


1.064 


0.940 


70 


1.222 


0.366 


21 


0.358 


2.790 


0.384 


2.605 


1.07 I 


0.934 


69 


1.204 


0.384 


22 


0.375 


2.669 


0.404 


2.475 


1.079 


0.927 


68 


1.187 


0.401 


23 


0.391 


2.559 


0.424 


2.356 


1.086 


0.921 


67 


1. 169 


0.419 


24 


0.407 


2.459 


0.445 


2.246 


1.095 


0.914 


66 


1.152 


0.436 


25 


0.423 


2.360 


0.466 


2.145 


1. 103 


0.906 


65 


1. 134 


0.454 


26 


0.438 


2.281 


0.488 


2.050 


1. 113 


0.899 


64 


J. 117 


0.471 


27 


0.454 


2.203 


0.510 


1.963 


1. 122 


0.891 


63 


1.099 


0.489 


28 


0.469 


2.T30 


0.532 


i.88r 


I. 133 


0.883 


62 


1.082 


0.506 


29 


0.485 


2.063 


0.554 


1.804 


.^.143 


0.875 


61 


1.064 


0.523 


30 


0.500 


2.000 


0.577 


1.732 


I-I55 


0.866 


60 


1.047 


0.541 


31 


0.5^5 


1.942 


0.601 


1.664 


1. 167 


0.857 


59 


1.030 


0.558 


32 


0-530 


1.887 


0.625 


1.600 


1. 179 


0.848 


58 


1. 012 


0.576 


33 


0.545 


1.836 


0.649 


1.540 


1. 192 


0.839 


57 


0.995 


0.593 


34 


0.559 


1.788 


0.675 


1.483 


1.206 


0.829 


56 


0.977 


0.61 1 


35 


0.574 


1.743 


0.700 


1.428 


1. 221 


0.819 


55 


0.960 


0.628 


36 


0.588 


1. 701 


0.727 


1.376 


1.236 


0.809 


54 


0.942 


0.646 


37 


0.602 


J. 662 


0.754 


1.327 


1.252 


0.799 


53 


0.925 


0.663 


38 


0.616 


1.624 


0.781 


1.280 


1.269 


0.788 


52 


0.908 


0.681 


39 


0.629 


1.589 


0.810 


1.235 


1.287 


0.777 


51 


0.890 


0.698 


40 


0.643 


1.556 


0.839 


1. 192 


1.305 


0.766 


50 


0.873 


0.716 


41 


0.656 


1.524 


0.869 


1. 150 


T.325 


0.755 


49 


0.855 


0.733 


42 


0.669 


1.494 


0.900 


I. Ill 


1.346 


0.743 


48 


0.838 


0.750 


43 


0.682 


1.466 


0.933 


1.072 


1.367 


0.731 


47 


0.820 


0.768 


44 


0.695 


1.440 


0.966 


1.036 


1.390 


0.719 


46 


0.803 


0.785 


45 


0.707 


1.414 


1. 000 


1. 000 


1. 414 


0.707 


45 

degr. 


0.785 




cos 


sec 


cot 


tan 


CSC 


sin 


arc 



TRIGONOMETRIC RATIOS (Natural) 


; including 


y ' ' arc," by which is meant 


the"7r- 1 


measure " or 


'circular measure 


of the angle; e.g., arc IOC = 


1.7453293. 


= moi7t. 1 


arc 


degr. 



sin 


CSC 


tan 


cot 


sec 


COS 






o.ooo 


0.000 


inf. 


0.000 


inf. 


1. 000 


1.000 


90 


1.571 


0.017 


I 


0.017 


57.3 


0.017 


57.3 


T.OOO 


1. 000 


89 


1.553 


0.035 


2 


0.035 


28.7 


0.035 


28.6 


1. 001 


0.999 


88 


1.536 


0.052 


3 


0.052 


I9.I 


0.052 


19.I 


1. 001 


0.999 


87 


1. 518 


0.070 


4 


0.070 


14.3 


0.070 


14.3 


1.002 


0.998 


86 


1.501 


0.087 


5 


0.087 


II-5 


0.087 


1 1.4 


1.004 


0.996 


85 


1.484 


0.105 


6 


0.105 


9.6 


0.105 


9.5 


1.006 


0.995 


84 


1.466 


0. I 2 2 


7 


0.122 


8.2 


0.123 


8.1 


1.008 


0.993 


^3 


T.449 


0.139 


8 


0.139 


7.2 


O.141 


7.1 


l.OIO 


0.990 


82 


1.432 


0.157 


9 


0.156 


6.4 


0.158 


6.3 


i.oi2 


0.988 


81 


1.414 


0.174 


10 


0.174 


5.8 


0.176 


5.7 


1.015 


0.985 


80 


1.396 


0.192 


11 


0.19 1 


5.24 


0.194 


5.14 


I.0I9 


0.982 


79 


1.379 


0.209 


12 


0.208 


4.8 1 


0.213 


4.70 


1.022 


0.978 


78 


1.361 


0.227 


13 


0.225 


4.45 


0.231 


4.33 


1.026 


0.974 


77 


1.344 


0.244 


14 


0.242 


4.13 


0.249 


4.01 


I.03I 


0.970 


76 


1.326 


0.262 


15 


0.259 


3.86 


0.268 


3.73 


1.035 


0.966 


75 


1.309 


0.279 


16 


0.276 


3-63 


0.287 


3-49 


1.040 


0.961 


74 


1.291 


0.297 


17 


0.292 


3.42 


0.306 


3.27 


1.046 


0.956 


73 


1.274 


0.314 


18 


0.309 


3.24 


0.325 


3.08 


1. 051 


0.95 y 


72 


1.25.7 


0.332 


19 


0.326 


3.07 


0.344 


2.90 


1.058 


0.946 


71 


1.239 


0.349 


20 


0.342 


2.92 


0.364 


2.75 


1.064 


0.940 


70 


1.222 


0.366 


21 


0.358 


2.790 


0.384 


2.605 


1. 071 


0.934 


69 


1.204 


0.384 


22 


0.375 


2.669 


0.404 


2.475 


1.079 


0.927 


68 


1.187 


0.401 


23 


0.391 


2-559 


0.424 


2.356 


1.086 


0.921 


67 


1.169 


0.419 


24 


0.407 


2.459 


0.445 


2.246 


1.095 


0.914 


66 


1.152 


0.436 


25 


0.423 


2.366 


0.466 


2.145 


1.103 


0,906 


65 


1. 134 


0.454 


26 


0.438 


2.281 


0.488 


2.050 


1.113 


0.899 


64 


1.117 


0.471 


27 


0.454 


2.203 


0.510 


1.963 


1. 122 


0.891 


63 


1.099 


0.489 


28 


0.469 


2.130 


0.532 


1.881 


^'^33 


0.883 


62 


1.082 


0.506 


29 


0.485 


2.063 


0.554 


1.804 


I.I43 


0.875 


61 


1.064 


0.523 


30 


0.500 


2.000 


0.577 


1.732 


I.I55 


0.866 


60 


1.047 


0.541 


31 


0.515 


1.942 


0.601 


1.664 


1.167 


0.857 


59 


1.030 


0.558 


32 


0.530 


1.887 


0.625 


1.600 


1. 179 


0.848 


58 


1.012 


0.576 


33 


0.545 


1.836 


0.649 


1.540 


1.192 


0.839 


57 


0.995 


0.59^ 


34 


0.559 


1.788 


0.675 


1.483 


1.206 


0.829 


56 


0.977 


O.61I 


35 


0.574 


1.743 


0.700 


1.428 


1.221 


0.819 


55 


0.960 


0.628 


36 


0.588 


1. 701 


0.727 


1.376 


1.236 


0.809 


54 


0.942 


0.646 


37 


0.602 


1.662 


0.754 


1.327 


1.252 


0.799 


53 


0.925 


0.663 


3^ 


0.616 


i.624 


0,781 


1.280 


1.269 


0.788 


52 


0.908 


0.681 


39 


0.629 


1.589 


0.810 


1.235 


1.287 


0.777 


51 


0.890 


0.698 


40 


0.643 


1.556 


0.839 


1.192 


1.305 


0.766 


50 


0.873 


0.716 41 1 


0.656 


1.524 


0,869 


1.150 


1.325 


0.755 


49 


0.855 


0.733 


42 


0.669 


1.494 


0.900 


1.111 


1.346 


0.743 


48 


0.838 


0.750 


43 


0.682 


1.466 


0.933 


1.072 


^'3^7 


0.731 


47 


0.820 


0.768 


44 


0.695 


1.440 


0.966 


1.036 


1,390 


0.719 


46 


0.803. 


0.785 


45 


0.707 


1:414 


1.000 


1,060 


1.414 


0.707 


45 

degr. 


0.785 




cos 


sec 


cot 


tan 


CSC 


sin 


arc ■ 



LOGAEITHMS (BEIGGS^). 


N 





1 


2 


3 


4 


5 





7 


8 


9 


Bif. 


lO 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


42 


11 


0414 


0453 


0492 


0531 


0569 


0607 


0645 


0682 


0719 


0755 


38 


13 


0792 


0828 


0864 


0899 


0934 


0969 


1004 


1038 


1072 


1106 


35 


13 


1139 


1173 


1206 


1239 


1271 


1303 


1335 


1367 


1399 


1430 


32 


14 


1461 


1492 


1523 


1553 


1584 


1614 


1644 


1673 


1703 


1732 


30 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


28 


16 


2041 


2068 


2095 


2122 


2148 


2175 


2201 


2227 


2253 


2279 


26 


17 


2304 


2330 


2355 


2380 


2405 


2430 


2455 


2480 


2504 


2529 


25 


18 


2553 


2577 


2601 


2625 


2648 


2672 


2695 


2718 


2742 


2765 


24 


19 


2788 


2810 


2833 


2856 


2878 


2900 


2923 


2945 


2967 


2989 


22 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


21 


21 


3222 


3243 


3263 


3284 


3304 


3324 


3345 


3365 


3385 


3404 


20 


22 


3424 


3444 


3464 


3483 


3502 


3522 


3541 


3560 


3579 


3598 


19 


23 


3617 


3636 


3655 


3674 


3692 


3711 


3729 


3747 


3766 


3784 


19 


24 


3802 


3820 


3838 


3856 


3874 


3892 


3909 


3927 


3945 


3962 


18 


25 


3979 


3997 


4014 


4031 


4048 


4065 


4082 


4099 


4116 


4133 


17 


26 


4150 


4166 


4183 


4200 


4216 


4232 


4249 


4265 


4281 


4298 


16 


27 


4314 


4330 


4346 


4362 


4378 


4393 


4409 


4425 


4440 


4456 


16 


28 


4472 


4487 


4502 


4518 


4533 


4548 


4564 


4579 


4594 


4609 


15 


29 


4624 


4639 


4654 


4669 


4683 


4698 


4713 


4728 


4742 


4757 


15 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


14 


31 


4914 


4928 


4942 


4955 


4969 


4983 


4997 


5011 


5024 


5038 


14 


32 


5051 


5065 


5079 


5092 


5105 


5119 


5132 


5145 


5159 


5172 


13 


33 


5185 


5198 


5211 


5224 


5237 


5250 


5263 


5276 


5289 


5302 


13 


34 


5315 


5328 


5340 


5353 


5366 


5378 


5391 


5403 


5416 


5428 


13 


35 


5441 


5453 


5465 


5478 


5490 


5502 


5514 


5527 


5539 


5551 


12 


36 


5563 


5575 


5587 


5599 


5611 


5623 


5635 


5647 


5658 


5670 


12 


37 


568a 


5694 


5705 


5717 


5729 


5740 


5752 


5763 


5775 


5786 


12 


38 


5798 


5809 


5821 


5832 


5843 


5855 


5866 


5877 


5888 


5899 


11 


39 


5911 


5922 


5933 


5944 


5955 


5966 


5977 


5988 


5999 


6010 


11 


40 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


11 


41 


6128 


6138 


6149 


6160 


6170 


6180 


6191 


6201 


6212 


6222 


10 


42 


6232 


6243 


6253 


6263 


6274 


6284 


6294 


6304 


6314 


6325 


10 


43 


6335 


6345 


6355 


6365 


6375 


6385 


6395 


6405 


6415 


6425 


10 


44 


6435 


6444 


6454 


6464 


6474 


6484 


6493 


6503 


6513 


6522 


10 


45 


6532 


6542 


6551 


6561 


6571 


6580 


6590 


6599 


6609 


6618 


10 


46 


6628 


6637 


6646 


6656 


6665 


6675 


6684 


6693 


6702 


6712 


9 


47 


6721 


6730 


6739 


6749 


6758 


6767 


6776 


6785 


6794 


6803 


9 


48 


6812 


6821 


6830 


6839 


6848 


6857 


6866 


6875 


6884 


6893 


9 


49 


6902 


6911 


6920 


6928 


6937 


6946 


6955 


6964 


6972 


6981 


9 


50 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


9 


51 


7076 


7084 


7093 


7101 


7110 


7118 


7126 


7135 


7143 


7152 


9 


52 


7160 


7168 


7177 


7185 


7193 


7202 


7210 


7218 


7226 


7235 


8 


53 


7243 


7251 


7259 


7267 


7275 


7284 


7292 


7300 


7308 


7316 


8 


54 


7324 


7332 


7340 


7348 


7356 


7364 


7372 


7380 


7388 


7396 


8 






N 


\ B. — Naperian log = 


= Brigg 


s'log 


X 2.302. 










Base of 


Naperian systt 


)m = e 


= 2.71828. 















LOGAKITHMS (BRIGGS'). 






1 


N 





1 


2 


3 


4 


5 


6 


7 


8 


9 


Dif. 


55 


7404 


7412 


7419 


7427 


7435 


7448 


7451 


7459 


7466 


7474 


8 


56 


7482 


7490 


7497 


7505 


7513 


7520 


7528 


7536 


7543 


7551 


8 


57 


7559 


7566 


7574 


7582 


7589 


7597 


7604 


7612 


7619 


7627 


8 


58 


7634 


7642 


7649 


7657 


7664 


7672 


7679 


7686 


7694 


7701 


7 


59 


7709 


7716 


7723 


7731 


7738 


7745 


7752 


7760 


7767 


7774 


7 


60 


7782 


7789 


7796 


7803 


7810 


7818 


7825 


7832 


7839 


7846 


7 


61 


7853 


7860 


7868 


7875 


7882 


7889 


7896 


7908 


7910 


7917 


7 


62 


7924 


7981 


7938 


7945 


7952 


7959 


7966 


7978 


7980 


7987 


7 


63 


7993 


8000 


8007 


8014 


8021 


8028 


8085 


8041 


8048 


8055 


7 


64 


8062 


8069 


8075 


8082 


8089 


8096 


8102 


8109 


8116 


8122 


7 


65 


8129 


8136 


8142 


8149 


8156 


8162 


8169 


8176 


8182 


8189 


7 


66 


8195 


8202 


8209 


8215 


8222 


8228 


8235 


8241 


8248 


8254 


7 


67 


8261 


8267 


8274 


8280 


8287 


8298 


8299 


8306 


8312 


8319 


6 


68 


8325 


8881 


8338 


8344 


8351 


8857 


8368 


8370 


8376 


8882 


6 


69 


8388 


8395 


8401 


8407 


8414 


8420 


8426 


8432 


8439 


8445 


6 


70 


8451 


8457 


8468 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


6 


71 


8513 


8519 


8525 


8581 


8537 


8548 


8549 


8555 


8561 


8567 


6 


72 


8578 


8579 


8585 


8591 


8597 


8603 


8609 


8615 


8621 


8627 


6 


73 


8633 


8689 


8645 


8651 


8657 


8668 


8669 


8675 


8681 


8686 


6 


74 


8692 


8698 


8704 


8710 


8716 


8722 


8727 


8738 


8739 


8745 


6 


75 


8751 


8756 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8802 


6 


76 


8808 


8814 


8820 


8825 


8831 


8837 


8842 


8848 


8854 


8859 


6 


77 


8865 


8871 


8876 


8882 


8887 


8893 


8899 


8904 


8910 


8915 


6 


78 


8921 


8927 


8932 


8938 


8948 


8949 


8954 


8960 


8965 


8971 


6 


79 


8976 


8982 


8987 


8993 


8998 


9004 


9009 


9015 


9020 


9025 


5 


80 


9031 


9036 


9042 


9047 


9058 


9058 


9063 


9069 


9074 


9079 


5 


81 


9085 


9090 


9096 


9101 


9106 


9112 


9117 


9122 


9128 


9183 


5 


82 


9138 


9143 


9149 


9154 


9159 


9165 


9170 


9175 


9180 


9186 


5 


88 


9191 


9196 


9201 


9206 


9212 


9217 


9222 


9227 


9282 


9238 


5 


84 


9243 


9248 


9258 


9258 


9263 


9269 


9274 


9279 


9284 


9289 


5 


85 


9294 


9299 


9304 


9309 


9815 


9820 


9825 


9330 


9335 


9340 


5 


86 


9345 


9850 


9355 


9360 


9365 


9870 


9375 


9380 


9385 


9390 


5 


87 


9895 


9400 


9405 


9410 


9415 


9420 


9425 


9430 


9485 


9440 


5 


88 


9445 


9450 


9455 


9460 


9465 


9469 


9474 


9479 


9484 


9489 


5 


89 


9494 


9499 


9504 


9509 


9513 


9518 


9523 


9528 


9588 


9538 


5 


90 


9542 


9547 


9552 


9557 


9562 


9566 


9571 


9576 


9581 


9586 


5 


91 


9590 


9595 


9600 


9605 


9609 


9614 


9619 


9624 


9628 


9638 


5 


92 


9688 


9648 


9647 


9652 


9657 


9661 


9666 


9671 


9675 


9680 


5 


93 


9685 


9689 


9694 


9699 


9708 


9708 


9718 


9717 


9722 


9727 


5 


94 


9731 


9736 


9741 


9745 


9750 


9754 


9759 


9768 


9768 


9773 


5 


95 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 


5 


96 


9823 


9827 


9882 


9886 


9841 


9845 


9850 


9854 


9859 


9863 


4 


97 


9868 


9872 


9877 


9881 


9886 


9890 


9894 


9899 


9903 


9908 


4 


98 


9912 


9917 


9921 


9926 


9980 


9934 


9989 


9948 


9948 


9952 


4 


99 


9956 


9961 


9965 


9969 


9974 


9978 


9983 


9987 


9991 


9996 


4 






K B.- 


Naperian log : 


= Briggs' log 


X 2.302. 










Base of 


Naperian Sys 


tern = 


e = 2.71828. 









